Abstract Nonsense

Crushing one theorem at a time

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 | , , , , , ,


  1. Why no love for section 19? (I’m working through the book to study for the math subject GRE.)

    Comment by Pat | June 7, 2010 | Reply

  2. n/m. Looking at the problems they pretty much seem to be the same. You working out of the first edition?

    Comment by Pat | June 7, 2010 | Reply

  3. Haha, yeah recheck the title, I just mislabeled it. And no, I am working out of the second edition.

    Also, is there really that much top. on the GREs?

    Comment by drexel28 | June 7, 2010 | Reply

  4. I don’t think there’s too much, which is why I only plan to work my way through the first three chapters. Afterwards will be a crash course in algebra (Fraleigh’s book which seems good). Keep it up, really cool what you’re doing!

    Comment by Pat | June 8, 2010 | Reply

  5. Thank you! I appreciate that someone actually read this. You’ll notice that this blog mostly used to be theorems, but I just really like doing problems. So, I figured over the summer I’d “try” to do all the problems in the “trifecta”: Topology-Munkres, Algebra-Dummit and Foote (over Herstein,yes), and Analysis-Rudin (baby Rudin). To be fair, I know all of these subjects fairly well. In fact, I’ve been through most of Munkres and Rudin’s book themselves. That said, it’ll still be hard.

    P.S. I would not suggest Fraleigh. That book is kind of dumbed down and if you intend to pursue a graduate degree in mathematics (as the GRE reference suggests) it won’t do.

    Comment by drexel28 | June 8, 2010 | Reply

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