Abstract Nonsense

Crushing one theorem at a time

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.

1.

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

2.

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

Solution: This is simple.

3.

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.

4.

Problem:

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.

Proof:

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.

5.

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

6.

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

7.

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.

Solution:

\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.

8.

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.

9.

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

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May 20, 2010 - Posted by | Fun Problems, Munkres, Topology | , , , , , ,

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