Abstract Nonsense

Crushing one theorem at a time

Munkres Chapter 2 Section 19 (Part II)


Problem: Show that the axiom of choice (AOC) is equivalent to the statement that for any indexed family \left\{U_{\alpha}\right\}_{\alpha\in\mathcal{A}} of non-empty sets, with \mathcal{A}\ne\varnothing, the product \displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha\ne\varnothing

Proof: This is pretty immediate when one writes down the actual definition of the product, namely:

\displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha=\left\{\bold{x}:\mathcal{A}\to\bigcup_{\alpha\in\mathcal{A}}U_{\alpha}:\bold{x}(\alpha)\in U_\alpha,\text{ for every }\alpha\in\mathcal{A}\right\}

So, if one assumes the AOC then one must assume the existence of a choice function

\displaystyle c:\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}\to\bigcup_{\alpha\in\mathcal{A}},\text{ such that }c\left(U_\alpha\right)\in U_\alpha\text{ for all }\alpha\in\mathcal{A}

So, then if we consider \left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}\overset{\text{def.}}{=}\Omega as just a class of sets, the fact that we have indexed them implies there exists a surjective “indexing function$


where clearly since we have already indexed out set we have that i:\alpha\mapsto U_\alpha. So, consider

c\circ i:\mathcal{A}\to\bigcup_{\alpha\in\mathcal{A}}U_\alpha

This is clearly a well-defined mapping and \left(c\circ i\right)(\alpha)=c\left(U_\alpha\right)\in U_\alpha and thus

\displaystyle c\circ i\in\prod_{\alpha\in\mathcal{A}}U_\alpha

from where it follows that \displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha\ne\varnothing

Conversely, let \Omega be a class of sets and let i:\mathcal{A}\to\Omega be an indexing function. We may then index \Omega by \Omega=\left\{U_{\alpha}\right\}_{\alpha\in\mathcal{A}}. Then, by assumption

\displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha\ne\varnothing

Thus there exists some

\displaystyle \bold{x}\in\prod_{\alpha\in\mathcal{A}}U_\alpha

Such that

\displaystyle \bold{x}:\mathcal{A}\to\bigcup_{\alpha\in\mathcal{A}},\text{ such that }\bold{x}(\alpha)\in U_\alpha

Thus, we have that

\displaystyle \bold{x}\circ i^{-1}:\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}\to\bigcup_{\alpha\in\mathcal{A}}

is a well-defined mapping with

\displaystyle \bold{x}\left(U_\alpha\right)\in U_\alpha

For each \alpha\in\mathcal{A}. It follows that we have produced a choice function for \Omega and the conclusion follows. \blacksquare

Remark: We have assumed the existence of a bijective indexing function i:\mathcal{A}\to\Omega, but this is either A) a matter for descriptive set theory or B) obvious since \text{id}:\Omega\to\Omega satisfies the conditions. This depends on your level of rigor.


Problem: Let A be a set; let \left\{X_\alpha\right\}_{\alpha\in\mathcal{A}} be an indexed family of spaces; and let \left\{f_\alpha\right\}_{\alpha\in\mathcal{A}} be an indexed family of functions f_\alpha:A\to X_\alpha

a) Prove there is a unique coarsest topology \mathfrak{J} on A relative to whish each of the functions f_\alpha is continuous.

b) Let

\mathcal{S}_\beta=\left\{f_\beta^{-1}\left(U_\beta\right):U_\beta\text{ is ope}\text{n in}X_\beta\right\}

and let \displaystyle \mathcal{S}=\bigcup_{\alpha\in\mathcal{A}}. Prove that \mathcal{S} is a subbasis for \mathfrak{J}.

c) Show that the map g:Y\to A is continuous relative to \mathfrak{J} if and only if each map f_\alpha\circ g:Y\to X_\alpha is continuous.

d) Let \displaystyle f:A\to\prod_{\alpha\in\mathcal{A}}X_\alpha be defined by the equation


Let Z denote the subspace of f\left(A\right) of the product space \displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha. Prove taht the image under f of each element of \mathfrak{J} is an open set in Z.


a) We first prove a lemma

Lemma: Let \mathfrak{J} be a topology on A, then all the mappings f_\alpha:A\to X_\alpha are continuous if and only if \mathcal{S}\subseteq\mathfrak{J} where \mathcal{S} is defined in part b).

Proof:Suppose that all the mappings f_\alpha:A\to X_\alpha are continuous. Then, given any open set U_\alpha\in X_\alpha we have that f_\alpha is continuous and so f_\alpha^{-1}\left(U_\alpha\right) is open and thus f_{\alpha}^{-1}\left(U_\alpha\right)\in\mathfrak{J} from where it follows that \mathcal{S}\subseteq\mathfrak{J}.

Conversely, suppose that \mathcal{S}\subseteq\mathfrak{J}. It suffices to prove that f_\alpha:A\to X_\alpha  for a fixed but arbitrary \alpha\in\mathcal{A}. So, to do this let U be open in X_\alpha then f_{\alpha}^{-1}\left(U\right)\in\mathfrak{J} and thus by assumption f_\alpha^{-1}\left(U\right)\in\mathfrak{J}; but this precisely says that f_\alpha^{-1}\left(U\right) is open in A. By prior comment the conclusion follows. \blacksquare

So, let

\mathcal{C}=\left\{\mathfrak{T}:\mathfrak{J}\text{ is a topology on }A\text{ and }\mathcal{S}\subseteq\mathfrak{J}\right\}

and let

\displaystyle \mathfrak{J}=\bigcap_{\mathfrak{I}\in\mathcal{C}}\mathfrak{T}

By previous problem \mathfrak{J} is in fact a topology on A, and by our lemma we also know that all the mappings f_\alpha:A\to X_\alpha are continuous since \mathcal{S}\subseteq\mathfrak{J}. To see that it’s the coarsest such topology let \mathfrak{U} be a topology for which all of the f_\alpha:A\to X_\alpha are continuous. Then, by the other part of our lemma we know that \mathcal{S}\subseteq\mathfrak{U} and thus \mathfrak{U}\in\mathcal{C}. So,

\displaystyle \mathfrak{J}=\bigcap_{\mathfrak{T}\in\mathcal{C}}\mathfrak{T}\subseteq\mathfrak{U}

And thus \mathfrak{J} is coarser than \mathfrak{U}.

The uniqueness is immediate.

b) It follows from the previous problem that we must merely show that \mathcal{S} is a subbasis for the topology \mathfrak{J}. The conclusion will follow from the following lemma (which was actually an earlier problem, but we reprove here for referential reasons):

Lemma: Let X be a set and \Omega be a subbasis for a topology on X. Then, the topology generated by \Omega equals the intersection of all topologies which contain \Omega.

Proof: Let

\mathcal{C}\left\{\mathfrak{T}:\mathfrak{T}\text{ is a topology on }X\text{ and }\Omega\subseteq\mathfrak{T}\right\}


\displaystyle \mathfrak{J}=\bigcap_{\mathfrak{T}\in\mathcal{C}}\mathfrak{T}

Also, let \mathfrak{G} be the topology generated by the subbasis \Omega.

Clearly since \Omega\subseteq\mathfrak{G} we have that \mathfrak{J}\subseteq\mathfrak{G}.

Conversely, let U\in\mathfrak{G}. Then, by definition to show that U\in\mathfrak{J} it suffices to show that U\in\mathfrak{T} for a fixed but arbitrary \mathfrak{T}\in\mathcal{C}. To do this we first note that by definition that

\displaystyle U=\bigcup_{\alpha\in\mathcal{A}}U_\alpha

where each

U_\alpha=O_1\cap\cdots\cap O_{m_\alpha}

for some O_1,\cdots,O_{m_\alpha}\in\Omega. Now, if \mathfrak{T}\in\mathcal{C} we know (since \Omega\subseteq\mathfrak{T}) that O_1,\cdots,O_{m_\alpha}\in\mathfrak{T} and thus

O_1\cap\cdots\cap O_{m_\alpha}=U_\alpha\in\mathfrak{T}

for each \alpha\in\mathcal{A}. It follows that U is the union of sets in \mathfrak{T} and thus U\in\mathfrak{T}. It follows from previous comment that \mathfrak{G}\subseteq\mathfrak{J}.

The conclusion follows. \blacksquare

The actual problem follows immediately from this.

c) So, let g:Y\to A be some mapping and suppose that f_\alpha\circ g:Y\to X_\alpha is continuous for each \alpha\in\mathcal{A}. Then, given a subbasic open set U in A we have that

U=f_{\alpha_1}^{-1}\left(U_{\alpha_1}\right)\cap\cdots\cap f_{\alpha_n}^{-1}\left(U_{\alpha_n}\right)

for some \alpha_1,\cdots,\alpha_n and for some open sets U_{\alpha_1},\cdots,U_{\alpha_n} in X_{\alpha_1},\cdots,X_{\alpha_n} respectively. Thus g^{-1}(U) may be written as

\displaystyle g^{-1}\left(\bigcup_{j=1}^{n}f_{\alpha_j}^{-1}\left(U_{\alpha_j}\right)\right)=\bigcup_{j=1}^{n}g^{-1}\left(f_{\alpha_j}^{-1}\left(U_{\alpha_j}\right)\right)=\bigcup_{j=1}^{n}\left(f_{\alpha_j}\circ g\right)^{-1}\left(U_{\alpha_j}\right)

but since each f_{\alpha_j}\circ g:Y\to X_{\alpha_j} we see that g^{-1}\left(U\right) is the finite union of open sets in Y and thus open in Y. It follows that g is continuous.

Conversely, suppose that g is continuous then f_\alpha\circ g:Y\to X_{\alpha} is continuous since it’s the composition of continuous maps.

d) First note that

\displaystyle f^{-1}\left(\prod_{\alpha\in\mathcal{A}}U_\alpha\right)=\bigcap_{\alpha\in\mathcal{A}}f\alpha^{-1}\left(U_\alpha\right)

from where it follows that the initial topology under the class of maps \{f_\alpha\} on A is the same as the initial topology given by the single map f. So, in general we note that if X is given the initial topology determined by f:X\to Y then given an open set f^{-1}(U) in X we have that f\left(f^{-1}(U)\right)=U\cap f(X) which is open in the subspace f(X).

June 9, 2010 Posted by | Fun Problems, Munkres, Topology | , , , , , , , | 1 Comment

Munkres Chapter 2 Section 19 (Part I)


Problem: Suppose that for each \alpha\in\mathcal{A} the topology on X_\alpha is given by a basis \mathfrak{B}_\alpha. The collection of all sets of the form

\displaystyle \prod_{\alpha\in\mathcal{A}}B_\alpha

Such that B_\alpha\in\mathfrak{B}_\alpha is a basis for \displaystyle X=\prod_{\alpha\in\mathcal{A}}X_\alpha with the box topology, denote this collection by \Omega_B. Also, the collection \Omega_P of all sets of the form

\displaystyle \prod_{\alpha\in\mathcal{A}}B_\alpha

Where B_\alpha\in\mathfrak{B}_\alpha for finitely many \alpha and B_\alpha=X_\alpha otherwise is a basis for the product topology on X.

Proof: To prove the first part we let U\subseteq X be open. Then, by construction of the box topology for each (x_\alpha)_{\alpha\in\mathcal{A}}\overset{\text{def.}}{=}(x_\alpha)\in U we may find some \displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha such that U_\alpha is open in X_\alpha and \displaystyle (x_\alpha)\in \prod_{\alpha\in\mathcal{A}}U_\alpha\subseteq U. So, then for each x_\alpha we may find some B_\alpha\in\mathfrak{B}_\alpha such that x_\alpha\in B_\alpha\subseteq U_\alpha and thus

\displaystyle (x_\alpha)\in\prod_{\alpha\in\mathcal{A}}B_\alpha\subseteq\prod_{\alpha\in\mathcal{A}}U_\alpha\subseteq U

Noticing that \displaystyle \prod_{\alpha\in\mathcal{A}}B_\alpha\in\Omega_B and every element of \Omega_B is open finishes the argument.

Next, we let U\subseteq X be open with respect to the product topology. Once again for each (x_\alpha)\in U we may find some \displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha such that U_\alpha is open in X_\alpha for each \alpha\in\mathcal{A} and U_\alpha=X_\alpha for all but finitely many \alpha, call them \alpha_1,\cdots,\alpha_m. So, for each \alpha_k,\text{ }k=1,\cdots,m we may find some B_k\in\mathfrak{B}_k such that x\in B_k\subseteq U_k and so

\displaystyle (x_\alpha)\in\prod_{\alpha\in\mathcal{A}}V_\alpha\subseteq\prod_{\alpha\in\mathcal{A}}U_\alpha\subseteq U


\displaystyle V_\alpha=\begin{cases}B_k\quad\text{if}\quad \alpha=\alpha_k ,\text{ }k=1,\cdots,m\\ X_\alpha\quad\text{if}\quad x\ne \alpha_1,\cdots,\alpha_m\end{cases}

Noting that \displaystyle \prod_{\alpha\in\mathcal{A}}V_\alpha\in\Omega_P and \Omega_P is a collection of open subsets of X finishes the argument. \blacksquare


Problem: Let \left\{U_\alpha\right\}_{\alpha\in\mathcal{A}} be a collection of topological spaces such that U_\alpha is a subspace of X_\alpha for each \alpha\in\mathcal{A}. Then, \displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha=Y is a subspace of \displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha=X if both are given the product or box topology.

Proof: Let \mathfrak{J}_S,\mathfrak{J}_P denote the topologies Y inherits as a subspace of X and as a product space respectively. Note that \mathfrak{J}_S,\mathfrak{J}_P are generated by the bases \mathfrak{B}_S=\left\{B\cap Y:B\in\mathfrak{B}\right\} (where \mathfrak{B} is the basis on X with the product topology), and

\displaystyle \mathfrak{B}_P=\left\{\prod_{\alpha\in\mathcal{A}}O_\alpha:O_\alpha\text{ is open in }U_\alpha,\text{ and equals }U_\alpha\text{ for all but finitely many }\alpha\right\}

So, let (x_\alpha)\in B where B\in\mathfrak{B}_S then

\displaystyle B=Y\cap\prod_{\alpha\in\mathcal{A}}V_\alpha=\prod_{\alpha\in\mathcal{A}}\left(V_\alpha\cap U_\alpha\right)

Where V_\alpha is open in X_\alpha, and thus V_\alpha\cap Y is open in U_\alpha. Also, since V_\alpha=X_\alpha for all but finitely many \alpha it follows that V_\alpha\cap U_\alpha=U_\alpha for all but finitely many \alpha. And so B\in\mathfrak{B}_P. Similarly, if B\in\mathfrak{B}_P then

\displaystyle B=\prod_{\alpha\in\mathcal{A}}O_\alpha

Where O_\alpha is open in U_\alpha, but this means that O_\alpha=V_\alpha\cap U_\alpha for some open set V_\alpha in X_\alpha and so

\displaystyle B=\prod_{\alpha\in\mathcal{A}}O_\alpha=\prod_{\alpha\in\mathcal{A}}\left(V_\alpha\cap U_\alpha\right)=\prod_{\alpha\in\mathcal{A}}V_\alpha\cap Y\in\mathfrak{B}_S

From where it follows that \mathfrak{B}_S,\mathfrak{B}_P and thus \mathfrak{J}_S,\mathfrak{J}_P are equal.

The case for the box topology is completely analgous. \blacksquare


Problem: Let \left\{X_\alpha\right\}_{\alpha\in\mathcal{A}} be a collection of Hausdorff spaces, then \displaystyle X=\prod_{\alpha\in\mathcal{A}}X_\alpha is Hausdorff with either the box or product topologies

Proof: It suffices to prove this for the product topology since the box topology is finer.

So, let (x_\alpha),(y_\alpha)\in X be distinct. Then, x_\beta\ne y_\beta for some \beta\in\mathcal{A}. Now, since X_\beta is Hausdorff there exists disjoint open neighborhoods U,V of x_\beta,y_\beta respectively. So, \pi_\beta^{-1}(U),\pi_\beta^{-1}(V) are disjoint open neighborhoods of (x_\alpha),(y_\alpha) respectively. The conclusion follows. \blacksquare


Problem: Prove that \left(X_1\times\cdots\times X_{n-1}\right)\times X_n\overset{\text{def.}}{=}X\approx X_1\times\cdots\times X_n\overset{\text{def.}}{=}Y.

Proof: Define

\displaystyle \varphi:X\to Y:((x_1,\cdots,x_{n-1}),x_n)\mapsto (x_1,\cdots,x_n)

Clearly this is continuous since \pi_{\beta}\circ\varphi=\pi_\beta


Problem: One of the implications holds for theorem 19.6 even in the box topology, which is it?

Proof: If \displaystyle f:A\to\prod_{\alpha\in\mathcal{A}}X_\alpha where the latter is given the box topology then we have that each \pi_\alpha is continuous and thus so is each \pi_\alpha\circ f:A\to X_\alpha. \blacksquare#


Problem: Let \left\{\bold{x}_n\right\}_{n\in\mathbb{N}} be a sequence of points in the product space \displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha=X. Prove that \left\{\bold{x}_n\right\}_{n\in\mathbb{N}} converges to \bold{x} if and only if the sequences \left\{\pi_\alpha(\bold{x}_n)\right\}_{n\in\mathbb{N}} coverge to \pi_\alpha(\bold{x}) for each \alpha\in\mathcal{A}. Is this fact true if one uses the box topology?

Proof: Suppose that U is a neighborhood of \pi_{\alpha}(\bold{x}) such that


is infinite. Notice then that if \pi_{\alpha}(\bold{x}_n)\in K that \bold{x}_n\notin\pi_{\alpha}^{-1}\left(U\right) from where it follows that \pi_{\alpha}^{-1}\left(U\right) is a neighborhood of \bold{x} which does not contain all but finitely many values of \left\{\bold{x}_n:n\in\mathbb{N}\right\} contradicting the fact that \bold{x}_n\to\bold{x} in X.

Conversely, suppose that \pi_{\alpha}(\bold{x}_n)\to\pi_{\alpha}(\bold{x}) for each \alpha\in\mathcal{A}  and let \displaystyle \prod_{\alpha\in\mathcal{A}}U_{\alpha} be a basic open neighborhood of \bold{x}. Then, letting \alpha_1,\cdots,\alpha_m be the finitely many indices such that U_{\alpha_k}\ne X_k. Since each \pi_{\alpha_k}(\bold{x}_n)\to\pi_{\alpha_k}(\bold{x}) there exists some n_\ell\in\mathbb{N} such that n_\ell\leqslant n\implies \pi_{\alpha_k}(\bold{x}_k)\in U_k. So, let N=\max\{n_1,\cdots,n_k\}. Now, note that if \displaystyle \bold{x}_n\notin\prod_{\alpha\in\mathcal{A}}U_\alpha then \pi_{\alpha}(\bold{x}_n)\notin U_\alpha for some \alpha\in\mathcal{A}. But, since clearly \pi_{\alpha}(\bold{x}_n)\in X_\alpha we must have that \pi_{\alpha}(\bold{x}_n)\notin U_{\alpha_1}\cup\cdots\cup U_{\alpha_m} and thus n\leqslant N. It follows that for every N\leqslant n we have that \displaystyle \bold{x}_n\in\prod_{\alpha\in\mathcal{A}}U_\alpha. Then, since every neighborhood of \bold{x} contains a basic open neighborhood and the above basic open neighborhood was arbitrary the conclusion follows.

Clearly the first part holds in the box topology since there was no explicit mention of the product topology or it’s defining characteristics. That said, the second does not hold. Consider \displaystyle \bold{x}_n=\prod_{m\in\mathbb{N}}\left\{\frac{1}{n}\right\}. Clearly each coordinate converges to zero, but \displaystyle \prod_{m\in\mathbb{N}}\left(\frac{-1}{m},\frac{1}{m}\right)=U is a neighborhood of \bold{0} in the product topology. But, if one claimed that for every n\geqslant N (for some N\in\mathbb{N} that \bold{x}_n\in U they’d be wrong. To see this merely note \displaystyle \frac{1}{N}\notin\left(\frac{-1}{N+1},\frac{1}{N+1}\right) and so \pi_{N+1}\left(\bold{x}_N\right)\notin\pi_{N+1}(U) and thus \bold{x}_{N}\notin U.


Problem: Let \mathbb{R}^{\infty} be the subset of \mathbb{R}^{\omega} consisting of all eventually zero sequences. What is \overline{\mathbb{R}^{\infty}} in the box and product topology?

Proof: We claim that in the product topology \overline{\mathbb{R}^{\infty}}=\mathbb{R}^{\omega}. To see this let \displaystyle \prod_{n\in\mathbb{N}}U_n be a basic non-empty open set in \mathbb{R}^{\omega} with the product topology. Since we are working with the product topology we know there are finitely many indices n_1,\cdots,n_m such that U_{n_k}\ne \mathbb{R}. So, for each n_1,\cdots,n_m select some x_{n_k}\in U_{n_k} and consider (x_n)_{n\in\mathbb{N}} where

\displaystyle \begin{cases} x_{n_k}\quad\text{if}\quad n=n_1,\cdots,n_m \\ 0 \text{ } \quad\text{if}\quad \text{otherwise}\end{cases}

Clearly then \displaystyle (x_n)_{n\in\mathbb{N}}\in \prod_{n\in\mathbb{N}}U_n\cap\mathbb{R}^{\infty} and thus every non-empty open set in \mathbb{R}^{\omega} intersects \mathbb{R}^{\infty} and the conclusion follows.

Now, we claim that with the box topology that \overline{\mathbb{R}^{\infty}}=\mathbb{R}^{\infty}. To see this let (x_n)_{n\in\mathbb{N}}\notin\mathbb{R}^{\infty}. Then, there exists some subsequence \{x_{\varphi(n)}\} of the sequence \{x_n\} which is non-zero. For each \varphi(n) form an interval I_{\varphi(n)} such that 0\notin I_{\varphi(n)}. Then, consider

\displaystyle \prod_{n\in\mathbb{N}}U_n,\text{ }U_n=\begin{cases} I_{\varphi(m)}\quad\text{if}\quad n=\varphi(m)\\ \mathbb{R}\quad\text{ }\quad\text{if}\quad \text{otherwise}\end{cases}

Clearly then \displaystyle \prod_{n\in\mathbb{N}}U_n is a neighborhood of (x_n)_{n\in\mathbb{N}} and since each clearly has an infinite subsequence of non-zero values it is disjoint from \mathbb{R}^{\infty}. It follows that in \mathbb{R}^{\omega} with the box topology that \mathbb{R}^{\infty} is closed and thus \overline{\mathbb{R}^{\infty}}=\mathbb{R}^{\infty} as desired. \blacksquare


Problem: Given sequences (a_n)_{n\in\mathbb{N}} and (b_n)_{n\in\mathbb{N}} of real numbers with a_n>0 define \varphi:\mathbb{R}^{\omega}\to\mathbb{R}^{\omega} by the equation

\varphi:(x_n)_{n\in\mathbb{N}}\mapsto \left(a_n x_n+b_n\right)_{n\in\mathbb{N}}

Show that if \mathbb{R}^{\omega} is given the product topology that \varphi is a homeomorphism. What happens if \mathbb{R}^{\omega} is given the box topology?

Proof: Let us first prove that \varphi is a bijection. To do this we prove something more general…

Lemma: Let \left\{X_\alpha\right\}_{\alpha\in\mathcal{A}} and \left\{Y_\alpha\right\}_{\alpha\in\mathcal{A}} be two classes of untopologized sets and \left\{f_\alpha\right\}_{\alpha\in\mathcal{A}} a collection of bijections f_\alpha:X_\alpha\to Y_\alpha. Then, if

\displaystyle \prod_{\alpha\in\mathcal{A}}f_\alpha\overset{\text{def.}}{=}\varphi:\prod_{\alpha\in\mathcal{A}}X_\alpha\to\prod_{\alpha\in\mathcal{A}}Y_\alpha:(x_\alpha)_{\alpha\in\mathcal{A}}\mapsto\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}

we have that \varphi is a bijection.

Proof: To prove injectivity we note that if




And by definition of an \alpha-tuple this implies that


for each \alpha\in\mathcal{A}. But, since each f_\alpha:X_\alpha\to Y_\alpha is injective it follows that


For each \alpha\in\mathcal{A}. Thus,


as desired.

To prove surjectivity we let \displaystyle (y_\alpha)_{\alpha\in\mathcal{A}}\in\prod_{\alpha\in\mathcal{A}}Y_\alpha be arbitrary. We then note that for each fixed \alpha\in\mathcal{A} we have there is some x_\alpha\in X_\alpha such that f_\alpha(x_\alpha)=y_\alpha. So, if \displaystyle (x_\alpha)_{\alpha\in\mathcal{A}}\in\prod_{\alpha\in\mathcal{A}}X_\alpha is the corresponding \alpha-tuple of these values we have that


from where surjectivity follows. Combining these two shows that \varphi is indeed a bijection. \blacksquare

Lemma: Let \left\{X_\alpha\right\}_{\alpha\in\mathcal{A}} and \left\{Y_\alpha\right\}_{\alpha\in\mathcal{A}} be two classes of non-empty topological spaces and \left\{f_\alpha\right\}_{\alpha\in\mathcal{A}} a corresponding class of continuous functions such that f_\alpha:X_\alpha\to Y_\alpha. Then, if \displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha and \displaystyle \prod_{\alpha\in\mathcal{A}}Y_\alpha are given the product topologies the mapping

\displaystyle \prod_{\alpha\in\mathcal{A}}f_\alpha\overset{\text{def.}}{=}\varphi:\prod_{\alpha\in\mathcal{A}}X_\alpha\to\prod_{\alpha\in\mathcal{A}}Y_\alpha:(x_\alpha)_{\alpha\in\mathcal{A}}\mapsto\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}

is continuous.

Proof: Since the codomain is a product space it suffices to show that

\displaystyle \pi_\beta\circ\varphi:\prod_{\alpha\in\mathcal{A}}X_\alpha\to Y_{\beta}

is continuous for each \beta\in\mathcal{A}. We claim though that the diagram

commutes where \pi^Y_\beta and \pi^X_\beta denote the canonical projections from \displaystyle \prod_{\alpha\in\mathcal{A}}Y_\alpha and \displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha to Y_\beta and X_\beta respectively. To see this we merely note that

\displaystyle \pi^Y_\beta\left(\varphi\left((x_\alpha)_{\alpha\in\mathcal{A}}\right)\right)=\pi^Y_\beta\left(\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}\right)=f_{\beta}(x_\beta)



which confirms the commutativity of the diagram. But, the conclusion follows since f_\beta\circ\pi_\beta is the composition of two continuous maps (the projection being continuous since \displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha is a product space).

The lemma follows by previous comment. \blacksquare

We come to our last lemma before the actual conclusion of the problem.

Lemma: Let \left\{X_\alpha\right\}_{\alpha\in\mathcal{A}} and \left\{Y_\alpha\right\}_{\alpha\in\mathcal{A}} be two classes of non-empty topological spaces and \left\{f_\alpha\right\}_{\alpha\in\mathcal{A}} a set of homeomorphisms with f_\alpha:X_\alpha\to Y_\alpha. Then,

\displaystyle \prod_{\alpha\in\mathcal{A}}f_\alpha\overset{\text{def.}}{=}\varphi:\prod_{\alpha\in\mathcal{A}}X_\alpha\to\prod_{\alpha\in\mathcal{A}}Y_\alpha:(x_\alpha)_{\alpha\in\mathcal{A}}\mapsto\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}

is a homeomorphism if \displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha and \displaystyle \prod_{\alpha\in\mathcal{A}}Y_\alpha are given the product topology.

Proof: Our last two lemmas show that \varphi is bijective and continuous. To prove that it’s inverse is continuous we note that

\displaystyle \left(\left(\prod_{\alpha\in\mathcal{A}}f_{\alpha}^{-1}\right)\circ\varphi\right)\left((x_\alpha)_{\alpha\in\mathcal{A}}\right)=\prod_{\alpha\in\mathcal{A}}\left\{f_\alpha^{-1}\left(f_\alpha(x_\alpha)\right)\right\}=(x_\alpha)_{\alpha\in\mathcal{A}}

And similarly for the other side. Thus,

\displaystyle \varphi^{-1}=\prod_{\alpha\in\mathcal{A}}f_{\alpha}^{-1}

Which is continuous since each f_{\alpha}^{-1}:Y_\alpha\to X_\alpha is continuous and appealing to our last lemma again. Thus, \varphi is a bi-continuous bijection, or in other words a homeomorphism. The conclusion follows. \blacksquare

Thus, getting back to the actual problem we note that if we denote T_n:\mathbb{R}\to\mathbb{R}:x\mapsto a_nx+b_n that each T_n is a homeomorphism. Thus, since it is easy to see that

\displaystyle \varphi=\prod_{n\in\mathbb{N}}T_n

we may conclude by our last lemma (since we are assuming that we are giving \mathbb{R}^{\omega} in both the domain and codomain the product topology) that \varphi is a homeomorphism.

This is also continuous if we give \mathbb{R}^{\omega} the box topology. To see this we merely need to note that \displaystyle \left(\prod_{\alpha\in\mathcal{A}}f_\alpha\right)^{-1}\left(\prod_{\alpha\in\mathcal{A}}U_\alpha\right)=\prod_{\alpha\in\mathcal{A}}f_\alpha^{-1}\left(U_\alpha\right) and thus if all of the U_\alpha are open then so are (since each f_\alpha is continuous) is each f_\alpha^{-1}(U_\alpha) and thus the inverse image of a basic open set is the unrestricted product of open sets and thus basic open. A similar argument works for the inverse function. \blacksquare

June 7, 2010 Posted by | Fun Problems, Munkres, Topology, Uncategorized | , , , , , , | 5 Comments

Munkres Chapter 2 Section 18


Problem: Show that the normal \varepsilon-\delta formulation of continuity is equivalent to the open set version.

Proof: Suppose that \left(\mathcal{M},d\right),\left(\mathcal{N},d'\right) are metric spaces and for every \varepsilon>0 and every f(x)\in\mathcal{N} there exists some \delta>0 such that f\left(B_{\delta}(x)\right)\subseteq B_{\varepsilon}(f(x)). Then, given an open set U\subseteq \mathcal{N} we have that f^{-1}(U). To see this let x\in f^{-1}(U) then f(x)\in U and since U is open by hypothesis there exists some open ball B_\varepsilon(f(x)) such that B_{\varepsilon}(f(x))\subseteq U and thus by assumption of \varepsilon-\delta continuity there is some \delta>0 such that \displaystyle f\left(B_{\delta}(x)\right)\subseteq B_{\varepsilon}(f(x))\subseteq U and so B_{\delta}\subseteq f^{-1}(U) and thus x is an interior point of f^{-1}(U).

Conversely, suppose that the preimage of an open set is always open and let f(x)\in\mathcal{N} and \varepsilon>0 be given. Clearly B_{\varepsilon}(f(x)) is open and thus f^{-1}\left(B_{\varepsilon}(f(x))\right) is open. So, since x\in f^{-1}\left(B_{\varepsilon}(f(x))\right) there exists some \delta>0 such that B_{\delta}(x)\subseteq f^{-1}\left(B_{\varepsilon}(f(x))\right) and so

f\left(B_{\delta}(x)\right)\subseteq f\left(f^{-1}\left(B_{\varepsilon}(f(x))\right)\right)\subseteq B_{\varepsilon}(f(x))



Problem: Suppose that f:X\to Y is continuous. If x is a limit point of the subset A of X, is it necessarily true that f(x) is a limit point of f(A)?

Proof: No. Consider (-1,0)\cup(0,1) with the suspace topology inherited from \mathbb{R} with the usual topology. Define

f:(-1,0)\cup(0,1)\to D:x\mapsto\begin{cases}0\quad\text{if}\quad x\in(-1,0)\\ 1\quad\text{if}\quad x\in(0,1)\end{cases}

This is clearly continuous since f^{-1}(\{1\})=(-1,0) and f^{-1}(\{1\})=(0,1) which are obviously open. But, notice that \frac{-1}{2} is a limit point for (-1,0) since given a neighborhood N of \frac{-1}{2} we must have that there is some (a,b)\cap \left((-1,0)\cup(0,1)\right)\cap \subseteq N which contains it. But, f\left(\frac{-1}{2}\right)=\{0\} is not a limit point for f\left((-1,0)\right)=\{0\} since that set has no limit points. \blacksquare


Problem: Let X and X' denote a singlet set in the two topologies \mathfrak{J} and \mathfrak{J}' respectively. Let \text{id}:X'\to X be the identity function. Show that

a) \text{id} is continuous if and only if \mathfrak{J}' is finer than \mathfrak{J}

b) \text{id} is a homeomorphism if and only if \mathfrak{J}=\mathfrak{J}'


a) Assume that \text{id} is continuous then given U\in\mathfrak{J} we have that \text{id}^{-1}(U)=U\in\mathfrak{J}'. Conversely, if \mathfrak{J}' is finer than \mathfrak{J} we have that given U\in\mathfrak{J} that \text{id}^{-1}(U)=U\in\mathfrak{J}'

b) If \text{id} is a homeomorphism we see that both it and \text{id}^{-1}=\text{id}:X\to X' are continuous and so mimicking the last argument we see that \mathfrak{J}\subseteq\mathfrak{J}' and \mathfrak{J}'\subseteq\mathfrak{J}. Conversely, if \mathfrak{J}=\mathfrak{J}' then we now that

U\in\mathfrak{J}\text{ iff }U\in\mathfrak{J}' or equivalently that U\text{ is open in }X\text{ iff  }\text{id}(U)=U\text{ is open in }X'

which defines the homeomorphic property. \blacksquare


Problem: Given x_0\in X and y_0\in Y show that the maps f:X\to X\times Y and g:X\times Y\to Y given by f:x\mapsto (x,y_0) and g:y\mapsto (x_0,y) are topological embeddings.

Proof: Clearly f and g are continuous since the projection functions are the identity and constant functions. They are clearly injective for if, for example, f(x)=(x,y_0)=(x',y_0)=f(x') then by definition of an ordered pair we must have that x=x'.  Lastly, the inverse function is continuous since f^{-1}:X\times \{y_0\}\to X:(x,y_0)\mapsto x is the restriction of the projection to X\times\{y_0\}. The same is true for g. \blacksquare


Problem: Show that with the usual subspace topology [0,1]\approx[a,b] and (0,1)\approx(a,b).

Proof: Define f:[0,1]\to[a,b]:x\mapsto (b-a)+a and g:(0,1)\to(a,b):x\mapsto (b-a)+a. These are easily both proven to be homeomorphisms. \blacksquare


Problem: Find a function f:\mathbb{R}\to\mathbb{R} which is continuous at precisely one point.

Proof: Define

f:\mathbb{R}\to\mathbb{R}:\begin{cases}x\quad\text{if}\quad x\in\mathbb{Q}\\ 0\quad\text{if}\quad x\notin\mathbb{Q}\end{cases}

Suppose that f is continuous at x_0, then choosing sequences \{q_n\}_{n\in\mathbb{N}},\{i_n\}_{n\in\mathbb{N}} of rational and irrationals numbers respectively both converging to x_0. We see by the limit formulation of metric space continuity that

x_0=\lim\text{ }q_n=\lim\text{ }f(q_n)=f(x_0)=\lim\text{ }f(i_n)=\lim\text{ }0=0

And so if f were to be continuous anywhere it would have to be at 0. To show that it is in fact continuous at 0 we let \varepsilon>0 be given then choosing \delta=\varepsilon we see that |x|<\delta\implies |f(x)|\leqslant |x|<\delta=\varepsilon from where the conclusion follows since this implies that \displaystyle \lim_{x\to 0}f(x)=0=f(0). \blacksquare



a) Suppose that f:\mathbb{R}\to\mathbb{R} is “continuous from the right”, that is, \displaystyle \lim_{x\to a^+}f(x)=f(a) for each a\in\mathbb{R}. Show that f is continuous when considered as a function from \mathbb{R}_\ell to \mathbb{R}.

b) Can you conjecture what kind of functions f:\mathbb{R}\to\mathbb{R} are continuous when considered as maps as \mathbb{R}\to\mathbb{R}_\ell. As maps from \mathbb{R}_\ell to \mathbb{R}_\ell?


a) Note that by the assumption that \displaystyle \lim_{x\to a^+}f(x)=f(a) we know that for every \varepsilon>0 there exists some \delta>0 such that 0\leqslant x-a<\delta implies that |f(x)-f(a)|<\varepsilon. So, let U\subseteq\mathbb{R} be open and let a\in f^{-1}(U). Then, f(a)\in U and since U is open we see that there is some \varepsilon>0 such that B_{\varepsilon}(f(a))\subseteq U. But, by assumption there exists some \delta>0 such that 0\leqslant x-a<\delta\implies f(x)\in B_{\varepsilon}(f(a)). But, \left\{x: 0\leqslant x-a<\delta\right\}=[a,a+\delta) and thus f\left([a,a+\delta)\right)\subseteq B_{\varepsilon}(f(a))\subseteq U and thus [a,a+\delta)\subseteq f^{-1}(U) and so a is an interior point for f^{-1}(U) from where it follows that f^{-1}(U) is open and thus f is continuous.

b) I’m not too sure, and not too concerned right now. My initial impression is that if f:\mathbb{R}\to\mathbb{R}_\ell is continuous then f^{-1}([a,b)) is open which should be hard to do. Etc.


Problem: Let Y be an ordered set in the order topology. Let f,g:X\to Y be continuous.

a) Show that the set \Omega=\left\{x\in X:f(x)\leqslant g(x)\right\} is closed in X

b) Let h:X\to Y:x\mapsto \max\{f(x),g(x)\}. Show that h is continuous.


a) Let x_0\notin\Omega then f(x_0)>g(x_0). Suppose first that there is no g(x_0)<\xi<f(x_0) and consider

f^{-1}\left(g(x_0),\infty)\right)\cap g^{-1}\left((-\infty,f(x_0)\right)=U

This is clearly open in X by the continuity of f,g and x_0 is contained in it. Now, to show that U\cap \Omega=\varnothing let z\in U then f(z)\in f\left(f^{-1}\left(g(x_0),\infty)\right)\cap g^{-1}\left(-\infty,f(x_0)\right)\right) which with simplification gives the important part that f(z)\in (g(x_0),\infty) and so f(z)>g(x_0) but since there is no \xi such that g(x_0)<\xi<f(x_0) this implies that f(z)\geqslant f(x_0). Similar analysis shows that g(z)\in (-\infty,f(x_0)) and since there is no \xi as was mentioned above this implies that g(z)\leqslant g(x_0). Thus, g(z)\leqslant g(x_0)<f(x_0)\leqslant f(z) and thus z\notin\Omega.

Now, suppose that there is some \xi such that g(x_0)<\xi<f(x_0) then letting V=f^{-1}(\xi,\infty)\cap g^{-1}(-\infty,\xi) we once again see that V is open and x_0\in V. Furthermore, a quick check shows that if z\in V that f(z)\in(\xi,\infty) and so f(z)>\xi and g(z)\in(-\infty,\xi) and so g(z)<\xi and so f(z)>g(z) so that z\notin\Omega. The conclusion follows

b) Let \Omega_f=\left\{x\in X:f(x)\geqslant g(x)\right\} and \Omega_g=\left\{x\in X:g(x)\geqslant f(x)\right\}. As was shown in a) both \Omega_f,\Omega_g are closed and thus define

f\sqcup g:X=\left(\Omega_f\cup\Omega_g\right)\to Y:x\mapsto\begin{cases}f(x)\quad\text{if}\quad x\in\Omega_f\\ g(x)\quad\text{if}\quad x\in\Omega_g\end{cases}

Notice that since f,g are both assumed continuous and f\mid_{\Omega_g\cap\Omega_f}=g\mid_{\Omega_f\cap\Omega_g} that we may conclude by the gluing lemma that f\sqcup g is in fact continuous. But, it is fairly easy to see that f\sqcup g=\max\{f(x),g(x)\} \blacksquare


Problem: Let \left\{U_\alpha\right\}_{\alpha\in\mathcal{A}} be a collection of subset of X; let \displaystyle X=\bigcup_{\alpha\in\mathcal{A}}U_\alpha. Let f:X\to Y and suppose that f\mid_{U_\alpha} is continuous for each \alpha\in\mathcal{A}

a) Show that if the collection \left\{U_\alpha\right\}_{\alpha\in\mathcal{A}} is finite each set U_\alpha is closed, then f is continuous.

b) Find an example where the collection \left\{U_\alpha\right\}_{\alpha\in\mathcal{A}} is countable and each U_\alpha is closed but f is not continuous.

c) An indexed family of sets \left\{U_\alpha\right\}_{\alpha\in\mathcal{A}} is said to be locally finite if each point of X has a neighborhood that intersects only finitely many elements of \left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}. Show that if the family \left\{U_\alpha\right\}_{\alpha\in\mathcal{A}} is locally finite and each U_\alpha is closed then f is continuous.$


a) This follows since if V\subseteq Y is closed then it is relatively easy to check that \displaystyle f^{-1}(V)=\bigcup_{\alpha\in\mathcal{A}}\left(f\mid_{U_\alpha}\right)^{-1}(V) but since each f\mid_{U_\alpha} is continuous we see that \left(f\mid_{U_\alpha}\right)^{-1}(V) is closed in U_\alpha. But, since each U_\alpha is closed in X it follows that each \left(f\mid_{U_\alpha}\right)^{-1}(V) is closed in X. Thus, f^{-1}(V) is the finite union of closed sets in X, and thus closed.

b) Give [0,1] the subspace topology inherited from \mathbb{R} with the usual topology and consider \left\{f_n\right\}_{n\in\mathbb{N}-\{1,2\}} with

f_n=\iota_{[0,1-\frac{1}{n}]}:\left[0,1-\tfrac{1}{n}\right]\to[0,1]:x\mapsto x

Clearly each f_n i

Lemma: Let Y be any topological space and \left\{V_\beta\right\}_{\beta\in\mathcal{B}} be a locally finite collection of subsets of Y. Then, \displaystyle \bigcup_{\beta\in\mathcal{B}}\overline{V_\beta}=\overline{\bigcup_{\beta\in\mathcal{B}}V_\beta}

Proof: The left hand inclusion is standard, so it suffices to prove the right inclusion. So, let \displaystyle x\in\overline{\bigcup_{\beta\in\mathcal{B}}V_\beta} since the collection of sets is locally finite there exists some neighborhood N of x such that it intersects only finitely many, say V_{\beta_1},\cdots,V_{\beta_n}, elements of the collection. So, suppose that x\notin \left(\overline{V_{\beta_1}}\cup\cdots\cup \overline{V_{\beta_n}}\right) then N\cap-\left(\overline{V_{\beta_1}}\cup\cdots\cup\overline{V_{\beta_n}}\right) is a neighborhood of x which does not intersect \displaystyle \bigcup_{\beta\in\mathcal{B}}V_\beta contradicting the assumption it is in the closure of that set. \blacksquare

Now, once again we let V\subseteq Y be closed and note that \displaystyle f^{-1}(V)=\bigcup_{\alpha\in\mathcal{A}}\left(f\mid_{U_\alpha}\right)^{-1}(V) and each \left(f\mid_{U_\alpha}\right)^{-1}(V) is closed in U_\alpha and since U_\alpha is closed in X we see that \left(f\mid_{U_\alpha}\right)^{-1}(V) is closed in X. So, noting that \left(f\mid_{U_\alpha}\right)^{-1}(V)\subseteq U_\alpha it is evident from the assumption that \left\{U_\alpha\right\}_{\alpha\in\mathcal{A}} is locally finite in X that so is \left\{\left(f\mid_{U_\alpha}\right)^{-1}(V)\right\}_{\alpha\in\mathcal{A}} and thus (for notational convenience) letting F_\alpha=\left(f\mid_{U_\alpha}\right)^{-1}(V) the above lemma implies that

\displaystyle \overline{f^{-1}(V)}=\overline{\bigcup_{\alpha\in\mathcal{A}}F_\alpha}=\bigcup_{\alpha\in\mathcal{A}}\overline{F_\alpha}=\bigcup_{\alpha\in\mathcal{A}}F_\alpha=f^{-1}(V)

From where it follows that the preimage of a closed set under f is closed. The conclusion follows. \blacksquare


Problem: Let f:A\to B and g:C\to D be continuous functions. Let us define a map f\times g:A\times C\to B\times D by the equation (f\times g)(a\times c)=f(a)\times g(c). Show that f\times g is continuous.

Proof: This follows from noting the two projections of f\times g are \pi_1\circ(f\times g):A\times B\to C:a\times b\mapsto f(a) and \pi_2\circ(f\times g):A\times B\to D:a\times b\mapsto f(b). But, both of these are continuous since \left(\pi_1(f\times g)\right)^{-1}(U)=f^{-1}(U)\times B. To see this we note that x\in f^{-1}(U)\times B if and only if x\in f^{-1}(U) which is true if and only if f(x)=\left(\pi_1\circ(f\times g)\right)(x)\in U or in other words x\in \left(\pi_1\circ(f\times g)\right)^{-1}(U). Using this we note that the preimage an open set in C will be the product of open sets by the continuity of f. It clearly follows both projections, and thus the function itself are continuous. \blacksquare


Problem: Let F:X\times Y\to Z. We say that F is continuous in eahc variable separately if for each y_0\in Y, the map h:X\to Z:x\mapsto F(x\times y_0( is continuous and for each x_0\in X the map j:Y\to Z:y\mapsto F(x_0\times y) is continuous. Show that if F is continuous then F is continuous in each variable separately.

Proof: If F is continuous then clearly it is continuous in each variable since if we denote by G_{y_0} the mapping G_{y_0}:X\to Z:x\mapsto F(x\times y_0) we see that G_{y_0}=H_{y_0}\circ(F\mid_{X\times\{y_0\}}) where H_{y_0}:X\to X\times Y:x\mapsto x\times y_0 but the RHS is the composition of continuous maps and thus continuous. A similar analysis holds for the other variable.


Problem: Let F:\mathbb{R}\times\mathbb{R}\to\mathbb{R} be given by

\displaystyle F(x\times y)=\begin{cases} \frac{xy}{x^2+y^2}&\mbox{if}\quad x\times y\ne 0\times 0\\ 0 &\mbox{if} \quad x\times y=0\times0\end{cases}

a) Show that F is continuous in each variable separately.

b) Compute g:\mathbb{R}\to\mathbb{R}:x\mapsto F(x\times x).

c) Show that F is not continuous


a) Clearly both F(x\times y_0) and F(x_0\times y) are continuous for x,y\ne 0 since they are the quotient of polynomials and the denominator is non-zero. Lastly, they are both continuous at x,y=0 since it is trivial to check that $

\displaystyle 0=F(0\times y_0)=F(x_0\times 0)=\lim_{x\to 0}F(x\times y_0)=\lim_{y\to 0}F(x_0\times y)

b) Evidently

\displaystyle g(x)=F(x\times x)=\begin{cases}\frac{1}{2}\quad\text{if}\quad x\times x\ne 0\\ 0\quad\text{if}\quad x\times x=0\end{cases}

c) This clearly proves that F(x\times y) is not continuous with \mathbb{R}^2 is not continuous since if \Delta is the diangonal we have that

\displaystyle \lim_{(x,y)\in\Delta\to (0,0)}F(x\times y)=\frac{1}{2}\ne F(0\times 0)

and so in particular

\displaystyle \lim _{(x,y)\to(0,0)}F(x\times y)\ne F(0\times 0)


Problem: Let A\subseteq X; let f:A\to X be continuous and let Y be Hausdorff. Prove that if f may be extended to a continuous function \overset{\sim}{f}:\overline{A}\to Y, then \overset{\sim}{f} is uniquely determined by f.

Proof: I am not too sure what this question is asking, but assuming it’s asking that if this extension existed it’s unique we can do this two ways

Way 1(fun way!):

Lemma: Let X be any topological space and Y a Hausdorff space. Suppose that \varphi,\psi:X\to Y are continuous and define A(\varphi,\psi)=\left\{x\in X:\varphi(x)=\psi(x)\right\}. Then, A(\varphi,\psi) is closed in X

Proof: Note that \varphi\oplus\psi:X\to Y\times Y:x\mapsto (\varphi(x),\psi(x)) is clearly continuous since \pi_1\circ(\varphi\oplus\psi)=\varphi and \pi_2\circ(\varphi\oplus\psi)=\psi. It is trivial then to check that \displaystyle A(\varphi,\psi)=\left(\varphi\oplus\psi\right)^{-1}(\Delta_Y) and since Y is Hausdorff we have that \Delta_Y\subseteq Y\times Y is closed and the conclusion follows. \blacksquare

From this we note that if \varphi,\psi agree on D\subseteq X such that \overline{D}=X we have that

X\supseteq A(\varphi,\psi)=\overline{A(\varphi,\psi)}\supseteq\overline{D}=X

From where it follows that A(\varphi,\psi)=X and so \varphi=\psi. So, thinking of \overline{A} as a subspace of X we see that \text{cl}_{\overline{A}}\text{ }A=Y\cap\text{cl}_{X}\text{ }A=\overline{A} and thus clearly A is dense in \overline{A}. So, the conclusion readily follows by noting that if \overset{\sim}{f_1},\overset{\sim}{f_2} are two continuous extensions then by definition A\left(\overset{\sim}{f_1},\overset{\sim}{f_2}\right)\supseteq A.

Way 2(unfun way): Let \overset{\sim}{f_1},\overset{\sim}{f_2} be two extensions of f and suppose there is some x\in\overline{A}-A(\varphi,\psi). Clearly x\notin A and thus x is a limit point of A. So, by assumption \overset{\sim}{f_1}(x)\ne\overset{\sim}{f_2}(x) and so using the Hausdorffness of Y we may find disjoint neighborhoods U,V of them respectively. Thus, \overset{\sim}{f_1}^{-1}(U),\overset{\sim}{f_2}^{-1}(V) are neighborhoods of x in X. Thus, \overset{\sim}{f_1}^{-1}(U)\cap\overset{\sim}{f_2}^{-1}(V) is a neighborhood of x. But, clearly there can be no y\in A\cap\left(\overset{\sim}{f_1}^{-1}(U)\cap\overset{\sim}{f_2}^{-1}(V)\right) otherwise \overset{\sim}{f_1}(y)=\overset{\sim}{f_2}(y)\in U\cap V. It follows that \overset{\sim}{f_1}^{-1}(U)\cap\overset{\sim}{f_2}^{-1}(V) is a neighborhood of x disjoint from A which contradicts the density of A in \overline{A}.  The conclusion follow. \blacksquare

May 28, 2010 Posted by | Fun Problems, Munkres, Topology, Uncategorized | , , , , , | 4 Comments

Munkres Chapter 2 Section 17

Theorem 17.11

Problem: Prove that every simply ordered set X with the order topology is Hausdorff. The product of two Hausdorff spaces is Hausdorff. A subspace of a Hausdorff space is Hausdorff.

Proof: Let X have the order topology and let a,b\in X be distinct and assume WLOG that a<b. If there does not exists a c\in X such that a<c<b then (-\infty,b),(a,\infty) are disjoint neighborhoods of a,b respectively. They are disjoint for to suppose that x\in(-\infty,b)\cap(a,\infty) would imply that a<x<b contradictory to our assumption. If there is some a<c<b then (-\infty,c),(c,\infty) are disjoint neighborhoods of a,b respectively.

Let X,Y be Hausdorff and (x,y),(x',y')\in X\times Y be distinct. Since (x,y)\ne(x',y') are distinct it follows we may assume WLOG that x\ne x'. But, since X is Hausdorff there exists disjoint neighborhoods U,V of x,y. So, consider U\times Y,V\times Y these are clearly neighborhoods of (x,y),(x',y') in X\times Y and to assume (u,v)\in U\times Y\cap V\times Y would imply \pi_1((u,v))-u\in\pi_1(U\times Y\cap V\times Y)\subseteq U\cap V.

Lastly, suppose that X is Hausdorff and let Y be a subspace of X. If x,y\in Y are distinct there exists disjoint neighborhoods U,V of them in X. Thus, U\cap Y,V\cap Y are disjoint neighborhoods of them in Y. \blacksquare


Problem: Let \mathcal{C} be a collection of subsets of the set X. Suppose that \varnothing,X\in\mathcal{C}, and that the finite union and arbitrary intersection of elements of \mathcal{C} are in \mathcal{C}. Prove that the collection \mathfrak{J}=\left\{X-C:C\in\mathcal{C}\right\} is a topology on X.

Proof: Clearly X=X-\varnothing and \varnothing=X-X are in \mathfrak{J}. Now, suppose that \left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}\subseteq\mathfrak{J} then \displaystyle X-\bigcup_{\alpha\in\mathcal{A}}U_\alpha=\bigcap_{\alpha\in\mathcal{A}}\left(X-U_\alpha\right) and since \left\{X-U_\alpha\right\}_{\alpha\in\mathcal{A}}\subseteq\mathcal{C} and \mathcal{C} is closed under arbitrary intersection it follows that \displaystyle \bigcap_{\alpha\in\mathcal{A}}\left(X-U_\alpha\right)=X-\bigcup_{\alpha\in\mathcal{A}}U_\alpha\in\mathcal{C} and thus \displaystyle \bigcup_{\alpha\in\mathcal{A}}U_\alpha\in\mathcal{C}. Lastly, suppose that \left\{U_1,\cdots,U_n\right\}\subseteq\mathcal{C} then \displaystyle X-\left(U_1\cap\cdots\cap U_n\right)=\left(X-U_1\right)\cup\cdots\cup\left(X-U_n\right) and since \left\{X-U_1,\cdots,X-U_n\right\}\subseteq\mathcal{C} and \mathcal{C} is closed under finite union it follows that X-\left(U_1\cap\cdots\cap U_n\right)\in\mathcal{C} and thus U_1\cap\cdots\cap U_n\in\mathfrak{J}. \blacksquare


Problem: Show that if A is a closed in Y and Y is closed in X that A is closed in X.

Proof: Since A is closed in Y and Y is a subspace of X we have that A=Y\cap G for some closed set G\subseteq X but since Y is closed it follows that A is the intersection of two closed sets in X and thus closed in X. \blacksquare


Problem: Show that if A is closed in X and B is closed in Y then A\times B is closed in X\times Y

Proof: This follows immediately from question 9. \blacksquare


Problem: Show that if U is open in X and A is closed in X, then U-A is open in X and A-U is closed in X.

Proof: This follows immediately from the fact that U-A=U\cap\left(X-A\right) and A-U=A\cap\left(X-U\right). \blacksquare


Problem: Let X be an ordered set in the order topology. Show that \overline{(a,b)}\subseteq[a,b]. Under what conditions does equality hold?

Proof: Let x\in(-\infty,a)\cup(b,\infty) then (-\infty,a)\cup(b,\infty) is a neighborhood of x disjoint from (a,b) and thus x\notin\overline{(a,b)}. Equality will hold when a,b are limit points of (a,b) or said otherwise whenever a<c<b we have that there are some $lated d,e$ such that a<d<c<e<b. \blacksquare


Problem: Let A,B and \left\{U_\alpha\right\}_{\alpha\in\mathcal{A}} denote subsets of X. Prove that if A\subseteq B then \overline{A}\subseteq\overline{B}, \overline{A\cup B}=\overline{A}\cup\overline{B}, and \displaystyle \overline{\bigcup_{\alpha\in\mathcal{A}}U_\alpha}\supseteq\bigcup_{\alpha\in\mathcal{A}}\overline{U_\alpha}. Give an example where this last inclusion is strict.

Proof: We choose to prove the second part first. Let x\notin\overline{A\cup B} then there is a neighborhood N of x such that N\cap (A\cup B)=(N\cap A)\cup (N\cap B)=\varnothing and thus x\notin \overline{A}\text{ and }x\notin\overline{B} and so x\notin\overline{A}\cup\overline{B}. Conversely, suppose that x\notin\overline{A}\cup\overline{B} then x\notin \overline{A} and x\notin\overline{B} and so there are neighborhoods N,G of x such that N\cap A=\varnothing,G\cap B=\varnothing clearly then N\cap G is a neighborhood of x such that (N\cap G)\cap(A\cup B)=\varnothing and thus x\notin\overline{A\cup B}.

Using this, if A\subseteq B then we have that \overline{B}=\overline{B\cup A}=\overline{B}\cup\overline{A} from where it follows that \overline{A}\subseteq\overline{B}.

Let \displaystyle x\notin\overline{\bigcup_{\alpha\in\mathcal{A}}U_\alpha} then there is a neighborhood N of x such that \displaystyle N\cap\bigcup_{\alpha\in\mathcal{A}}U_\alpha=\bigcup_{\alpha\in\mathcal{A}}\left(N\cap U_\alpha\right)=\varnothing and thus N\cap U_\alpha=\varnothing for every \alpha\in\mathcal{A} and thus x\notin\overline{U_\alpha} for every \alpha\in\mathcal{A} and so finally we may conclude that \displaystyle x\notin\bigcup_{\alpha\in\mathcal{A}}\overline{U_\alpha}.

To see when inclusion can be strict consider \mathbb{R} with the usual topology. Then,

\displaystyle \mathbb{R}=\overline{\bigcup{q\in\mathbb{Q}}\{q\}}\supsetneq\bigcup_{q\in\mathbb{Q}}\overline{\{q\}}=\bigcup_{q\in\mathbb{Q}}\{q\}=\mathbb{Q}



Problem: Criticize proof (see book).

Proof: There is no guarantee that the A_\alpha for which U intersected will b e the same A_\alpha that V will intersect if you pick another V. \blacksquare


Problem: Let A,B and \left\{U_\alpha\right\}_{\alpha\in\mathcal{A}} denote subsets of X. Determine whether the following equations hold, if any equality fails determine which inclusion holds.

a) \overline{A\cap B}=\overline{A}\cap\overline{B}

b) \displaystyle \overline{\bigcap_{\alpha\in\mathcal{A}}U_\alpha}=\bigcap_{\alpha\in\mathcal{A}}\overline{U_\alpha}

c) \overline{A-B}=\overline{A}-\overline{B}


a) Equation does not hold. Consider that \overline{(-1,0)\cap(0,1)}=\overline{\varnothing}=\varnothing but \overline{(-1,0)}\cap\overline{(0,1)}=[-1,0]\cap[0,1]=\{0\}. The \subseteq inclusion always holds.

b) This follows from a) that equality needn’t hold. Once again the \subseteq inclusion is true.

c) This need be true either \overline{\mathbb{R}-\mathbb{Q}}=\mathbb{R}\ne\overline{\mathbb{R}}-\overline{\mathbb{Q}}=\varnothing. The \subseteq inclusion holds. \blacksquare


Problem: Prove that if A\subseteq X,B\subseteq Y then \overline{A\times B}=\overline{A}\times\overline{B} in the product topology on X\times Y.

Proof: Let \displaystyle (x,y)\in\overline{A\times B}. Let U\times V be a basic open set in X\times Y which contains (x,y). Since x\in \overline{A},y\in \overline{B} we can choose some point x'\in U\cap A,y'\in V\cap B. Then, (x',y')\in U\times V\cap A\times B. It follows that \displaystyle (x,y)\in\overline{A\times B}

Conversely, let (x,y)\in\overline{A\times B}. Let U\subseteq X be any set such that x\in U. Since \pi_1^{-1}(U) is open in X\times Y it contains some point (x',y')\in A\times B. Then, x'\in U\cap A. It follows that x\in\overline{A}. A similar technique works for Y. \blacksquare

10., 11,. 12 Covered in theorem stated and proved at the beginning of the post.


Problem: Prove that X is Hausdorff if and only if the diagonal \Delta=\left\{(x,x):x\in X\right\} is closed in X\times X with the product topology.

Proof: Suppose that X is Hausdorff then given x\ne y we may find disjoint neighborhoods U,V of them. So, (x,y)\in U\times V and U\times V\cap \Delta=\varnothing since U\cap V=\varnothing. It follows that \Delta is closed.

Conversely, suppose \Delta is closed in X\times X and x,y\in X are distinct. Then, (x,y)\notin\Delta and so there exists a basic open neighborhood U\times V of (x,y) such that U\times V\cap \Delta=\varnothing and so U,V are neighborhoods of x,y in X which are disjoint. For, to suppose they were not disjoint would to assume that z\in U\cap V\implies (z,z)\in U\times V contradicting the assumption that U\times V\cap\Delta=\varnothing. \blacksquare


Problem: In the cofinite topology on \mathbb{R} to what point or points does the sequence \left\{\frac{1}{n}\right\}_{n\in\mathbb{N}} converge to?

Proof: It converges to every point of \mathbb{R}. To see this let x\in\mathbb{R} be arbitrary and let U be any neighborhood of it. Then, \mathbb{R}-U is finite and in particular \left(\mathbb{R}-U\right)\cap K is finite. If it’s empty we’re done, so assume not and let \frac{1}{n_0}=\min\left(\left(\mathbb{R}-U\right)\cap K\right) then clearly for all n\in\mathbb{N} such that n>n_0 we have that \frac{1}{n}<\frac{1}{n_0} and thus \frac{1}{n} is in U. The conclusion follows. \blacksquare


Problem: Prove that the T_1 axiom is equivalent to the condition that for each pair of points x,y\in X there are neighborhoods of each which doesn’t contain the other.

Proof: Suppose that X is T_1 then given distinct x,y\in X the sets X-\{y\},X-\{x\} are obviously neighborhoods of x,y respectively which don’t contain the other.

Conversely, suppose the opposite is true and let y\in X-\{x\} then there is a neighborhood U of it such that x\notin U\implies U\subseteq X-\{x\} and thus X-\{x\} is open and \{x\} is therefore closed. \blacksquare


Problem: Consider the five topologies on \mathbb{R} given in exercise 7 of section 13 (my section 2).

a) Deterime the closure of K under each of these topologies.

b) Which of these topologies are Hausdorff? Which are T_1


a) As a reminder the topologies are

\mathfrak{J}_1=\text{usual topology}

\mathfrak{J}_2=\text{topology on }\mathbb{R}_K

\mathfrak{J}_3=\text{cofinite topology}

\mathfrak{J}_4=\text{upper limit topology}

\mathfrak{J}_5=\text{left ray topology}

For the first one we easily see that \overline{K}=K\cup\{0\}.

For the second one we can see that \overline{K}=K. To see this note that \displaystyle \bigcup_{a<b}(a,b)-K=\mathbb{R}-K is open by definition and thus K being the complement of it is closed. Thus, \overline{K}=K

For the third one \overline{K}=\mathbb{R}. To see this note that we in a sense already proved this in 14, but for any x\in\mathbb{R} and any neighborhood N of it we have that N=\mathbb{R}-\{x_1,\cdots,x_n\} and thus if \displaystyle \frac{1}{n_0}=\min\left(\{x_1,\cdots,x_n\}\cap K\right) (assuming it’s non-empty) we see that n>n_0\implies \frac{1}{n}\notin\{x_1,\cdots,x_n\}\implies \frac{1}{n}\in N. Thus, given any point of \mathbb{R} and any neighborhood N of it we have that N\cap K is infinite, and thus clearly x\in\overline{K}. The conclusion follows from that.

For the fourth topology we note that \displaystyle \mathbb{R}-K=(-\infty,0]\cup\bigcup_{n\in\mathbb{N}}\left(\frac{1}{n+1},\frac{1}{n}\right)\cup(1,\infty) which is open in \mathbb{R} with the upper limit topology. Remember that \displaystyle (a,b)=\bigcup_{a<c<b}(a,c]

For the last one \overline{K}=[0,\infty). Clearly K\subseteq[0,\infty) and since [0,\infty) is closed and \overline{K} is the intersection of all closed supersets of K it follows that \overline{K}\subseteq[0,\infty). Now, suppose that x\notin[0,\infty) then x\in(-\infty,0) which is a neighborhood of x which doesn’t intersect K and thus x\notin\overline{K}. So, [0,\infty)\subseteq\overline{K}. The conclusion follows.


Lemma: If X is Hausdorff, then it is T_1

Proof: Let x\in X and y\in X-\{x\} by assumption there exists disjoint neighborhoods U,V of x,y respectively and so clearly y\in V\subseteq X-\{x\} and thus X-\{x\} is open and so \{x\} is closed. \blacksquare

\mathfrak{J}_1: Clearly the first topology is Hausdorff since it is an order topology and we proved in the initial post that all order topologies are Hausdorff.

Lemma: Let X be a set and \mathfrak{J},\mathfrak{J}' two topologies on X such that \mathfrak{J}' is finer than \mathfrak{J}. Then, if X is Hausdorff with the \mathfrak{J} topology it is Hausdorff with the \mathfrak{J}' topology.

Proof: Clearly if x,y\in X are distinct we may find disjoint neighborhoods U,V\in\mathfrak{J} of them in the topology given by \mathfrak{J} and thus the same neighborhoods work in consideration of the topology given by \mathfrak{J}'. \blacksquare

\mathfrak{J}: From this lemma it follows that \mathbb{R}_K having a finer topology than \mathbb{R} with the usual topology is Hausdorff

\mathfrak{J}_3: The cofinite topology is T_1 but not Hausdorff. To see that it’s T_1 it suffices to prove that \{x\} is closed for any x\in\mathbb{R}. But, this is trivial since \mathbb{R}-\{x\} being the complement of a finite set is open, thus \{x\} closed. But, it is not Hausdorff since it cannot support two disjoint, non-empty open sets. To see this suppose that U is open in \mathbb{R} with the cofinite topology then \mathbb{R}-U is finite and since a disjoint set V would have to be a subset of \mathbb{R}-U it follows that V is finite and thus it’s complement not finite. Thus, V is not open.

\mathfrak{J}_4: Once again this topology finer than that of \mathbb{R} with the usual topology since \displaystyle (a,b)=\bigcup_{c<b}(a,c]

\mathfrak{J}_5: This isn’t even T_1. To see this we must merely note that if U is any set containing 1 we must have that there is some basic open set (-\infty,\alpha),\text{ }\alpha>1 such that 1\in(-\infty,\alpha)\subseteq U, but this means that 0\in U. So, there does not exist a neighborhood of 1 which does not contain 0.


Problem: Consider the lower limit topology on \mathbb{R} and the topology given by the basis \mathcal{C}=\left\{[a,b):a<b,\text{ }a,b\in\mathbb{Q}\right\}. Determine the closure of the intervals A=(0,\sqrt{2}) and B=(\sqrt{2},3) in these two topologies.

Proof: We first prove more generally that if (a,b)\subseteq\mathbb{R}_\ell then \overline{(a,b)}=[a,b). To see this we first note that [a,b) is open since \mathbb{R}-[a,b)=(-\infty,a)\cup[b,\infty) which we claim is open. To see this we note that \displaystyle (-\infty,a)\cup[b,\infty)=\bigcup_{x<a}[x,a)\cup\bigcup_{y>b}[b,y). So, since \overline{(a,b)} is the intersection of all closed supersets of (a,b) and [a,b)\supseteq(a,b) is closed we see that \overline{(a,b)}\subseteq[a,b). So, we finish the argument by showing that \overline{(a,b)}\ne(a,b). To see this we show that a is a limit point for (a,b) from where the conclusion will follow. So, let N be any neighborhood of a, then we may find some basic open neighborhood [\alpha,\beta) such that a\in[\alpha,\beta)\subseteq N, but clearly [\alpha,\beta)\cap(a,b) contains infinitely many points from where it follows that a is a limit point of (a,b). From this we may conclude for \mathbb{R}_\ell that \overline{(0,\sqrt{2})}=[0,\sqrt{2}) and \overline{(\sqrt{2},3)}=[\sqrt{2},3)

We now claim that in the topology generated by \mathcal{C} that \overline{(0,\sqrt{2})}=[0,\sqrt{2}] and \overline{(\sqrt{2},3)}=[\sqrt{2},3). More generall, let us prove that in this topology

\overline{(a,b)}=\begin{cases}[a,b]\quad\text{if}\quad b\notin\mathbb{Q}\\ [a,b)\quad\text{if}\quad b\in\mathbb{Q}\end{cases}

Clearly if \alpha<a then choosing some rational number q<\alpha and some rational number \alpha<p<a then \alpha\in[q,p) and [q,p)\cap(a,b)=\varnothing so that \alpha\notin\overline{(a,b)}. Also, if \alpha>b then choosing some p\in\mathbb{Q} such that b<p<\alpha we see that \alpha\in[p,\alpha+1) and [p,\alpha+1)\cap(a,b)=\varnothing. Thus, the only possibilities for \overline{(a,b)} are (a,b),[a,b),[a,b]. But, just as before if N is any neighborhood a we may find some open basic neighborhood $latex [p,q)$ such that a\in[p,q)\subseteq N but clearly [p,q)\cap(a,b)\ne\varnothing and thus since N was arbitrary it follows that a\in\overline{(a,b)}.

So, now we split into the two cases. First assume that b\notin\mathbb{Q} then given any neighborhood N of b we may find some basic neighborhood [p,q)  such that b\in[p,q)\subseteq N, but since b\notin\mathbb{Q} we see that b\ne p from where it follows that [p,q)\cap(a,b)\ne\varnothing and thus since N was arbitrary it follows that b\in\overline{(a,b)} and thus \overline{(a,b)}=[a,b]. Now suppose that b\in\mathbb{Q}, then [b,b+1) is clearly a neighborhood of b that doesn’t intersect (a,b) and thus b\notin{(a,b)} so that \overline{(a,b)}=[a,b)

The conclusion follows. \blacksquare

Problem: If A\subseteq X, we define the boundary of A by \partial A=\overline{A}-\overline{X-A}.

a) Prove that A^\circ and \partial A are disjoint and \overline{A}=A^\circ\cup\partial A

b) Prove that \partial A=\varnothing i and only if A is both open and closed

c) Prove that U is open if and only if \partial U=\overline{U}-U

d) If U is open, is it true that U=\left(\overline{U}\right)^{\circ}?


a) Clearly if x\in A^{\circ} then there exists a neighborhood of x whose intersection with the complement of A is empty, thus x\notin\partial A. Now, let x\in\overline{A} now if there exists a neighborhood N of x such that N\subseteq A then x\in A^{\circ} and if not then every neighborhood of x contains points of X-A and since x\in D(A) it follows that it also contains points of A. Thus, either x\in A^{\circ} or x\in\partial A, thus x\in A^{\circ}\cup\partial A. Conversely, if x\in A^{\circ}\partial A then either there exists a neighborhood N of x such that N\subseteq A and thus x\in A.  Conversely, if x\in\partial A then for every neighborhood N of x we have that N\cap A,N\cap (X-A)=\ne\varnothing and in particular N\cap A\ne\varnothing and so x\in\overline{A}.

b) Suppose first that \partial A=\varnothing. If A=\varnothing we’re done, so assume not and let x\in A. Since x\notin\partial A there is a neighborhood N of x such that N\cap A\varnothing\text{ or }N\cap (X-A)=\varnothing but since x\in N\cap A it follows that N\subseteq A and thus A is open. Conversely, letting x\in X-A we see that x\notin\partial A and so by the same logic there exists a neighborhood N of x such that N\subseteq X-A and thus X-A is open and so A closed.

Conversely, suppose that A is both open and closed and suppose that \partial A\ne\varnothing. If x\in\partial A\cap A then for every neighborhood N of x we must have that N\cap (X-A)\ne\varnothing and thus x\notin A^{\circ}=A, which is a contradiction. Conversely, if x\in (X-A)\cap\partial A then for every neighborhood N of x we must have that N\cap A\ne\varnothing and thus x\notin (X-A)^{\circ}=X-A which is a contradiction.

c) Suppose that U is open and let x\in\partial U, then every neighborhood Nof x contains points of U by definition, but since x\notin U (it can’t be an interior point, thus x\notin U^{\circ}=U)  they must be points of U-\{x\} and thus x\in D(U). So, x\in \overline{U}-U. Conversely, if x\in \overline{U}-U then x must be a limit point of U which is not in U and thus every neighborhood x contains points of U and X-U so x\in\partial U

d) No, it is not true. With the usual topology on \mathbb{R} the set (-1,0)\cup(0,1) is open, but \left(\overline{(-1,0)\cup(0,1)}\right)^{\circ}=[-1,1]^\circ=(-1,1)

We choose to omit 18, 19, 21 on the grounds that they are monotonous and easy, monotonous and easy, and way, way to long.

May 22, 2010 Posted by | Fun Problems, Munkres, Topology, Uncategorized | , , , , | Leave a comment

Munkres Chapter 2 Section 1

This begins a substantial effort to complete all of (except the first chapter) the problems in James Munkre’s Topology I have chosen to start at chapter 2 considering the first chapter is nothing but prerequisites. Thus, without a minute to spare let us being.


Problem: Let X be a topological space, let A be a subset of X. Suppose that for each x\in A there is an open set U_x containing x such that U\subseteq A. Prove that A is open.

Proof: We claim that \displaystyle A=\bigcup_{x\in A}U_x=\overset{\text{def.}}{=}\Omega but this is obvious since for each x\in A we have that x\in U_x\subseteq\Omega. Conversely, since each U_x\subseteq A we have that the union of all of them is contained in A, namely \Omega\subseteq A. Thus, A is the union of open sets and thus open. \blacksquare


Problem: Compare the nine topologies on \{a,b,c\} given in example 1.

Solution: This is simple.


Problem: Show that given a set X if we denote \mathfrak{J} to be cocountable topology (a set is open if it’s complement is countable or the full space) that \left(X,\mathfrak{J}\right) is a topological space. Is it still a topological space if we let \mathfrak{J}=\left\{U\in\mathcal{P}(X):X-U\text{ is infinite, empty, or all of }X\right\}?

Proof: Clearly for the first part \varnothing,X\in\mathfrak{J}. Now, if \left\{U_\alpha\right\}_{\alpha\in\mathcal{A}} is a collection of open sets then we note that X-U_\alpha is finite and thus \displaystyle X-\bigcup_{\alpha\in\mathcal{A}}U_\alpha=\bigcap_{\alpha\in\mathcal{A}}\left(X-U_\alpha\right)\subseteq U_{\alpha_0} for any \alpha_0. Thus, \displaystyle X-\bigcup_{\alpha\in\mathcal{A}}U_\alpha is finite and thus in \mathfrak{J}. Now, if U_1,\cdots,U_n\in\mathfrak{J} we have that X-(U_1\cap\cdots\cap U_n)=(X-U_1)\cup\cdots\cup (X-U_n) and thus X-(U_1\cap\cdots\cap U_n) is the finite union of finite sets and thus finite, so U_1\cap\cdots\cap U_n\in\mathfrak{J}.

If we redefine the topology as described it is not necessarily a topology. For example, give \mathbb{N} that topology and note that \left\{\{n\}\right\}_{n\in\mathbb{N}-\{1\}} is a collection of elements of \mathfrak{J} but \displaystyle \bigcup_{n\in\mathbb{N}-\{1\}}\{n\}=\mathbb{N}-\{1\} whose complement is neither infinite, empty, or the full space. Thus, this isn’t a topology.



a) If \left\{\mathfrak{J}_\alpha\right\}_{\alpha\in\mathcal{A}} is a family of topologies on X, show that \displaystyle \bigcap_{\alpha\in\mathcal{A}}\mathfrak{J}_\alpha is a topology on X. Is \displaystyle \bigcup_{\alpha\in\mathcal{A}}\mathfrak{J}_\alpha?

b) Let \left\{\mathfrak{J}\right\}_\alpha be a family of topologies on X. Show that there is unique topology on X containing all the collections \mathfrak{J}_\alpha, and a unique largest topology contained in all of the \mathfrak{J}_\alpha.

c) If X=\{a,b,c\}, let \mathfrak{J}_1=\{\varnothing,X,\{a\},\{a,b\}\} and \mathfrak{J}_2=\left\{\varnothing,X,\{a\},\{b,c\}\right\}. Find the smallest topology containing \mathfrak{J}_1,\mathfrak{J}_2 and the larges topology contained in \mathfrak{J}_1,\mathfrak{J}_2.


a) Let \displaystyle \Omega\overset{\text{def.}}{=}\bigcap_{\alpha\in\mathcal{A}}\mathfrak{J}_\alpha and let \left\{U_\beta\right\}_{\beta\in\mathcal{B}}\subseteq\Omega be arbitrary. Then, by assumption we have that \left\{U_\beta\right\}_{\beta\in\mathcal{B}}\subseteq\mathfrak{J}_\alpha for every \alpha\in\mathcal{A} but since this was assumed to be a topology we have that for every \alpha\in\mathcal{A} that \displaystyle \bigcup_{\beta\in\mathcal{B}}U_\beta\subseteq\mathfrak{J}_\alpha and thus \displaystyle \bigcup_{\beta\in\mathcal{B}}U_\beta\in\Omega. Now, if \{U_1,\cdots,U_n\}\subseteq\Omega we must have that \{U_1,\cdots,U_n\}\subseteq\mathfrak{J}_\alpha for every \alpha and thus U_1\cap\cdots\cap U_n\in\mathfrak{J}_\alpha for every \alpha\in\mathcal{A} and thus U_1\cap\cdots\cap U_n\in\Omega. Thus, noting that for every \alpha\in\mathcal{A} we must have that \varnothing,X\in\mathfrak{J}_\alpha the conclusion follows.

No, the union of two topologies needn’t be a topology. Let \mathfrak{J}_1,\mathfrak{J}_2 be defined as in part c and note that \mathfrak{J}_1\cup\mathfrak{J}_2=\left\{\varnothing,X,\{a\},\{a,b\},\{b,c\}\right\} but \{a,b\}\cap\{b,c\}=\{c\}\notin\mathfrak{J}_1\cup\mathfrak{J}_2

b) This follows immediately from part a. For the first part let \Omega=\left\{\mathfrak{J}\in\text{Top }X:\mathfrak{J}_\alpha\subseteq\mathfrak{J},\text{ for all }\alpha\in\mathcal{A}\right\} (where \text{Top }X is the set of all topologies) and let \mathfrak{T}=\bigcap_{\mathfrak{J}\in\Omega}\mathfrak{J}, this clearly satisfies the conditions. For the second one merely take \displaystyle \bigcap_{\alpha\in\mathcal{A}}\mathfrak{J}_\alpha

c) We (as pointed out very poignantly in the previous problem) just take the union of the two topologies. So, we claim that the desired smallest topology is in fact \left\{X,\varnothing,\{a\},\{b\},\{a,b\},\{a,c\}\right\}. But this is just computation and we leave it to the reader. For the second part just intersect the two topologies.


Problem: Show that if \mathcal{A} is a base for a topology on X, then the topology generated by \mathcal{A} equals the intersection of all topologies on X which contain \mathcal{A}. Prove the same if \mathcal{A} is a subbase

Proof: Let \Omega be the intersection of all topologies on X which contain \mathcal{A} and \mathfrak{J}_g the topology generated by \mathcal{A}. Clearly \Omega\subseteq\mathfrak{J}_g since \mathfrak{J}_g is itself a topology on X containing \mathcal{A}. Conversely, let U\in\mathfrak{J}_g we show that U\in\mathfrak{J} where \mathfrak{J} is any topology on X containing X. But, this is obvious since \displaystyle U=\bigcup_{B\in\mathcal{B}}B for some \mathcal{B}\subseteq\mathcal{A} and thus U is the union of open sets in \mathfrak{J} and thus in \mathfrak{J}. The conclusion follows.

Next, let \Omega,\mathfrak{J}_g be above except now \mathcal{A} is a subbase. For the same reasons as above we have that \Omega\subseteq\mathfrak{J}_g. Conversely, for any topology \mathfrak{J} containing \mathcal{A} we have that if U\in\mathfrak{J}_g then \displaystyle U=\bigcup_{\beta\in\mathcal{B}}V_\beta where each V_\beta is the finite union of elements of \mathcal{A}. But, by construction it follows that each V_\beta is open (it is the finite intersection of open sets in \mathfrak{J})  and thus U is the union of open sets in \mathfrak{J} and thus V\in\mathfrak{J}. \blacksquare


Problem: Show that the topologies on \mathbb{R}_K and the Sorgenfrey line aren’t comparable

Proof: See the last part of the next problem


Problem: Consider the following topologies on\mathbb{R}:

\mathfrak{J}_1=\text{usual topology}

\mathfrak{J}_2=\text{topology on }\mathbb{R}_K

\mathfrak{J}_3=\text{cofinite topology}

\mathfrak{J}_4=\text{topology having the set set of all }(a,b]\text{ as a base}

\mathfrak{J}_5=\text{the topology having all sets }(-\infty,a)\text{ as a base}

For each determine which of the others contain it.


\mathfrak{J}_1\subseteq\mathfrak{J}_2:Clearly we have that \mathfrak{J}_1\subseteq\mathfrak{J}_2 since the defining open base for \mathfrak{J}_1 is contained entirely in \mathfrak{J}_2.

\mathfrak{J}_1\supseteq\mathfrak{J}_3: But, \mathfrak{J}_1\not\subseteq\mathfrak{J}_3 since (0,1)\in\mathfrak{J}_1 but \mathbb{R}-(0,1)\simeq\mathbb{R} and thus not in \mathfrak{J}_3. Now, to prove the inclusion indicated we know that for each open set U in the cofinite topology we have that U=\mathbb{R}-\{x_1,\cdots,x_n\} and so if we assume WLOG that x_1<\cdots<x_n then

\displaystyle U=\bigcup_{a<x_1}(a,x_1)\cup(x_1,x_2)\cup\cdots\cup(x_{n-1},x_n)\bigcup_{b>x_n}(x_n,b)

And thus U is open in the usual topology.

\mathfrak{J}_1\subseteq\mathfrak{J}_4: But, \mathfrak{J}_1\subseteq\mathfrak{J}_4. To see this it suffices to show that (a,b) is open in \mathfrak{J}_4 since this is a base for \mathfrak{J}_1. But, to see this we must merely note that \displaystyle (a,b)=\bigcup_{c<b}(a,c].

\mathfrak{J}_1\supseteq\mathfrak{J}_5: Lastly, \mathfrak{J}_1\supseteq\mathfrak{J}_5. To see this we must merely note that \displaystyle (-\infty,b)=\bigcup_{a<b}(a,b).

For \mathfrak{J}_2 the result is obvious except possibly how it relates to \mathfrak{J}_3. But, in fact they aren’t comparable. To see this we first show that (0,1] is not open in \mathbb{R}_K. To see this we show it can’t be written as the union of sets of the form (a,b) and (c,d)-K but it clearly suffices to do this for the latter sets. Now, to see that (0,1] can’t be written as the union of sets of the form (a,b) we recall from basic real number topology that (0,1] is not open (1 is not an interior point) and thus it can’t be written as the union of intervals or it would be open in the usual topology on \mathbb{R}. Also, consider (-1,1)-K.


Problem: Show that the countable collection \mathfrak{B}=\left\{(a,b):a<b,\text{ }a,b\in\mathbb{Q}\right\} is a base for the usual topology on \mathbb{R}

Proof: This follows from the density of \mathbb{Q}. It suffices to show that given any x\in\mathbb{R} and any (a,b)\supseteq\{x\} that there is some element (p,q)\in\mathfrak{B} such that x\in(p,q)\subseteq(a,b). But, from basic analysis we know there is some rational number q such that a<q<x and similarly there is some p\in\mathbb{Q} such that x<p<b. Thus, x\in(p,q)\subseteq(a,b). The conclusion follows. \blacksquare.


Problem: Show that the collection \mathcal{C}=\left\{[a,b):a<b,\text{ }a,b\in\mathbb{R}\right\} generates a different topology from the one on \mathbb{R}_\ell (the lower limit topology)$.

Proof: Clearly [\sqrt{2},3) is open in \mathbb{R}_\ell but we show that it can’t be written as the union of elements of \mathcal{C}. So, suppose that \displaystyle \bigcup_{\alpha\in\mathcal{A}}[a_\alpha,b_\alpha)=[\sqrt{2},3) where \left\{[a_\alpha,b_\alpha)\right\}_{\alpha\in\mathcal{A}}\subseteq\mathcal{C}. Then, there exists some [a_\alpha,b_\alpha) such that \sqrt{2}\in[a_\alpha,b_\alpha). Now, since [a_\alpha,b_\alpha)\subseteq[\sqrt{2},b_\alpha) we must have that a_\alpha\geqslant \sqrt{2} but a_\alpha\neq\sqrt{2} and thus a_\alpha>\sqrt{2} and thus \sqrt{2}\notin[a_\alpha,b_\alpha). Contradiction. \blacksquare

May 20, 2010 Posted by | Fun Problems, Munkres, Topology | , , , , , , | Leave a comment

Just For Fun(Rudin’s Topology Section) Part IV


Problem: Prove that ever closed set in a separable metric space \mathcal{M}  is the union of a (possibly empty) perfect set and a set which is a set which is countable.

Proof: It is easy to see that the above proof (proof 27.) holds in any space which is second countable, in particular for separable metric spaces. Thus, if E\subseteq\mathcal{M} is closed we surely have that E\subseteq \mathfrak{C}\cup\left(\left(\mathbb{R}-\mathfrak{C}\right)\cap E\right). Thus, E=\left(E\cap\mathfrak{C}\right)\cup\left(\left(\mathbb{R}-\mathfrak{C}\right)\cap E\right) but since \mathfrak{C}\subseteq D(E) and D(E)\subseteq E we have then that E=\mathfrak{C}\cup\left(\left(\mathbb{R}-\mathfrak{C}\right)\cap E\right) which is the union of a perfect and countable set respectively. \blacksquare


Problem: Prove that every open set U\subseteq\mathbb{R} may be written as the disjoint union of countably many open intervals.

Proof: We need to prove a quick lemma

Lemma: Let X be a topological space and let \left\{U_\alpha\right\}_{\alpha\in\mathcal{A}} be a class of connected subspace of X such that U_{\alpha_1}\cap U_{\alpha_2}\ne\varnothing,\text{ }\alpha_1,\alpha_2\in\mathcal{A} then \displaystyle \Lambda=\bigcup_{\alpha\in\mathcal{A}} is a connected subspace of X.

Proof: Suppose that \left(E\cap \Lambda\right)\cup\left(G\cap \Lambda\right)=\Lambda is a separation of \Lambda. We may assume WLOG that U_{\alpha_0}\cap E\ne\varnothing for some \alpha_0\in\mathcal{A}. So, now we see that U_{\alpha_0}\subseteq E\cap \Lambda otherwise E\cap U_{\alpha_0},G\cap U_{\alpha_0} would be non-empty disjoint subsets of U_{\alpha_0} whose union is U_{\alpha_0} contradicting that U_{\alpha_0} is connected. Thus, it easily follows that for any  U_\alpha we have that U_\alpha\cap U_{\alpha_0}\ne\varnothing so that U_{\alpha}\cap E\ne\varnothing and thus by a similar reasoning we see that U_\alpha\subseteq E\cap\Lambda. Thus, since \alpha was arbitrary it follows that \Lambda\subseteq E\cap\Lambda contradicting that G\cap\Lambda\ne\varnothing. \blacksquare

So, now for each x\in U define

\mathcal{C}(x)=\left\{V\subseteq U: V\text{ is open in }U,\text{ }V\text{ is connected },\text{ and }x\in V\right\}

And let \displaystyle C(x)=\bigcup_{V\in\mathcal{C}(x)}V and finally we prove that

\Omega=\left\{C(x):x\in U\right\}

is a countable class of disjoint open intervals whose union is U. The fact that each C(x) is an open interval follows since it is the union of open connected spaces with non-empty intersection (each element of \mathcal{C}(x) contains x) it is also an open connected subspace of \mathbb{R} (note that each element of \mathcal{C}(x) is open in U but since U is open it is also open in \mathbb{R}. But, it was proven in the book the only connected subspace of \mathbb{R} are intervals and thus C(x) is an interval for each C(x)\in\Omega.

Now, to see that they are disjoint we show that if C(x)\cap C(y)\ne\varnothing then C(x)=C(y) from where the conclusion will follow. So, to see this we first note that if C(x)\cap C(y) is non-empty then C(x)\cup C(y) is an open connected subspace of U containing both x and y and thus C(x)\cup C(y)\subseteq C(x) and C(x)\cup C(y)\subseteq C(y) but this implies that C(y)\subseteq C(x) and C(x)\subseteq C(y) respectively from where the conclusion follows.

Now, to see that \Omega is countable we notice that for each \Omega we have that C(x)\cap\mathbb{Q}\ne\varnothing and so if we let F\left(C(x)\right) be a fixed but arbitrary q\in C(x)\cap\mathbb{Q} then

F:\Omega\to\mathbb{Q}:C(x)\mapsto F\left(C(x)\right)

is an injection since the elements of \Omega are pairwise disjoint. The fact that \Omega is countable follows immediately.

Thus, \Omega is a countable collection of open intervals and

\displaystyle U=\coprod_{C(x)\in\Omega}C(x)

Thus, the conclusion follows. \blacksquare


Problem: Prove that if \displaystyle \mathbb{R}^n=\bigcup_{n\in\mathbb{N}}F_n where each F_n is a closed subset of \mathbb{R}^n then at least one F_n has non-empty interior.


Lemma: Let \mathcal{M} be a complete metric space and \left\{K_n\right\}_{n\in\mathbb{N}} a descending sequence of non-empty closed subsets of \mathcal{M} such that \text{diam }K_n\to 0. Then, \displaystyle \bigcap_{n\in\mathbb{N}}K_n contains one point.

Proof: Clearly \displaystyle \bigcap_{n\in\mathbb{N}}K_n does not contain more than one point since \text{diam }K_n\to 0. So, now for each n\in\mathbb{N} choose some x_n\in K_n and let P=\left\{x_n:n\in\mathbb{N}\right\}. Now, if P were somehow finite a similar argument to that used in the first case of problem 26. is applicable, so assume that P is infinite. But, since \text{diam }K_n\to 0 it is evident that \left\{x_n\right\}_{n\in\mathbb{N}} is a Cauchy sequence and thus by assumption it converges to some point x\in\mathcal{M}. We claim that \displaystyle x\in\bigcap_{n\in\mathbb{N}}K_n. To see this we note similarly to problem 26. that since P is infinite it is easy to see that x is a limit point of P and so if x\notin K_{n_0} for some n_0\in\mathbb{N} then \mathcal{M}-K_{n_0} is a neighborhood of x containing only finitely many points of P which clearly contradicts that it is a limit point. The conclusion follows. \blacksquare

So, now suppose each F_n had empty interior (i.e. nowhere dense) . Then, since \mathcal{M} is open and F_1 is nowhere dense there exists an open ball B_1 of radius less than one such that B_1\cap F_1=\varnothing. Let E_1  be the concentric closed ball of B_1 whose radius is half that of B_1. Since F_2 is nowhere dense E_1^{\circ} contains an open ball B_2 of radius less than one-half which is disjoint from F_2.  Let E_2 be the concentric closed ball of B_2 whose radius is one-half that of B_2. Since F_3 is nowhere dense we have that E_2^{\circ} contains an open ball B_3 of radius less than one-fourth which is disjoint from F_3. Let E_3 be the concentric closed ball of B_3 whose radius is half that of E_3. Continuing in this we get a descending sequence of non-empty closed subsets \left\{E_n\right\} for which \text{diam }E_n\to 0. Thus, by our lemma we have that there is some \displaystyle x\in\bigcap_{n\in\mathbb{N}}E_n. This point is clearly not in any of the F_n‘s from where the conclusion follows. \blacksquare

May 14, 2010 Posted by | Analysis, Fun Problems, Topology, Uncategorized | , , , , , , , , | Leave a comment