Abstract Nonsense

Crushing one theorem at a time

Munkres Chapter two Section 12 & 13: Topological Spaces and Bases


Point of post: This is the solutions to Munkres Chapter two Section 12 as the heading indicates.

1.

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

Proof: So, we know that for each x\in A there exists some open U_x such that x\in U_x\subseteq. Clearly

\displaystyle A=\bigcup_{x\in A}U_x

and so A is the union of open sets, and thus open.

2.

Problem: Consider the nine topologies on the set X=\{a,b,c\} indicated in Example 1 of section 12. Compare them.

Proof: This is incredibly laborious to type out, and really not that difficult, so we omit it.

3.

Problem: Show that the collection \mathfrak{J}_c given in example 4 of section 12 is a topology on the set X. Is the collection

\mathfrak{J}_\infty=\left\{U:X-U\text{ is infinite or empty or all of }X\right\}

a topology on X?

Proof: This \mathfrak{J}_c is just the cocountable topology. We can see that \varnothing\in\mathfrak{J}_c since it’s complement is all of X. Also, X\in\mathfrak{J}_c since X-X=\varnothing which is finite. Also, we notice that if \left\{U_{\alpha}\right\}_{\alpha\in\mathcal{A}} is a nonempty subset of \mathfrak{J}_c that

\displaystyle X-\bigcup_{\alpha\in\mathcal{A}}U_\alpha=\bigcap_{\alpha\in\mathcal{A}}\left(X-U_{\alpha}\right)\subseteq U_{\alpha_0},\text{ }\alpha_0\in\mathcal{A}

and so the complement of \displaystyle \bigcup_{\alpha\in\mathcal{A}}U_\alpha is a subset of a countable set and thus countable. Similarly, if U_1,\cdots,U_n\in\mathfrak{J}_c then

X-\left(U_1\cup\cdots\cup U_n\right)=\left(X-U_1\right)\cup\cdots\cup\left(X-U_n\right)

and so the complement of U_1\cap\cdots\cap U_n is the finite union of countable sets, and thus countable, so U_1\cap\cdots\cap U_n\in\mathfrak{J}_c. It follows that \mathfrak{J}_c really is a topology on X

Now, \mathfrak{J}_c is not a topology on any infinite set. To see this let X be infinite and let \varphi:X\to\mathbb{Z} be any surjection. Then, we can see that \varphi^{-1}\left(\{0\}\cup\mathbb{N}\right) and \varphi^{-1}\left(-\mathbb{N}\cup\{0\}\right) are in \mathbb{J}_{\infty} but their intersection is \{0\} which is not in \mathfrak{J}_\infty

4.

Problem:

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

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

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

Proof:

a) Clearly since \varnothing,X\in\mathfrak{J}_{\alpha} for every \alpha\in\mathcal{A}  we have that \varnothing,X\in\Omega. Now, if \left\{U_{\beta}\right\}_{\beta\in\mathcal{B}}\subseteq\Omega then \left\{U_{\beta}\right\}_{\beta\in\mathcal{B}}\subseteq \mathfrak{J}_\alpha for every \alpha\in\mathcal{A}. But, since each \mathfrak{J}_{\alpha} is a topology it follows that \displaystyle \bigcup_{\beta\in\mathcal{B}}U_\beta\in\mathfrak{J}_\alpha for every \alpha\in\mathcal{A} and so \displaystyle \bigcup_{\beta\in\mathcal{B}}U_\beta\in\Omega. Lastly, if U_1,\cdots,U_n\in\Omega we have that U_1,\cdots,U_n\in\mathfrak{J}_\alpha for every \alpha\in\mathcal{A}. But, since these are topologies this implies that U_1\cap\cdots\cap U_n\in\mathfrak{J}_\alpha for every \alpha\in\mathcal{A} and so U_1\cap\cdots\cap U_n\in\Omega

b) To show that there is a unique topology which contains all the \mathcal{J}_\alpha we define

\Gamma=\left\{\mathfrak{T}:\mathfrak{T}\text{ is a topology on }X\text{ and }\mathfrak{T}\supseteq\bigcup_{\alpha\in\mathcal{A}}\mathfrak{J}_\alpha\right\}

We first notice that \Gamma is not empty since 2^X\in\Gamma. So, let

\displaystyle \mathfrak{J}=\bigcap_{\mathfrak{T}\in\Gamma}\mathfrak{T}

Clearly then \mathfrak{J} is a topology which contains all the \mathfrak{J}_\alpha, and any other topology which contains them must contain \mathfrak{J}. More formally, \mathfrak{J} is a topology on X such that \mathfrak{J}_\alpha\subseteq\mathfrak{J},\text{ }\forall\alpha\in\mathcal{A}. Furthermore, given a topology \mathfrak{K} which contains every \mathfrak{J}_\alpha we see that \mathfrak{K}\in\Gamma so that \mathfrak{J}\subseteq\mathfrak{K}. Also, if \mathfrak{J}' were another smallest topology containing all the \mathfrak{J}_\alpha‘s that both the inclusions \mathfrak{J}\subseteq\mathfrak{J}' and \mathfrak{J}'\subseteq\mathfrak{J} and so \mathfrak{J}=\mathfrak{J}', from where it follows that the smallest such topology is unique.

Using the exact same logic, the smallest topology contained in all of the \mathfrak{J}_{\alpha}‘s is \displaystyle \mathfrak{J}=\bigcap_{\alpha\in\mathcal{A}}\mathfrak{J}_\alpha.

c) One can verify that \mathfrak{J}_1\cup\mathfrak{J}_2\cup\{b\} is the smallest topology containing both and \mathfrak{J}_1\cap\mathfrak{J}_2 is a topology, and thus the smallest contained in both.

5.

Problem: Show that if \mathfrak{B} is a basis for a topology on X, then the topology generated by \mathfrak{B} equals the intersection of all topologies on X that contain \mathfrak{B}. Show the same is true if \mathfrak{B} is a subbasis.

Proof:Let T(X) be the set of all topologies on X and let

\displaystyle \mathfrak{J}=\bigcap_{\mathfrak{I}\in T(X):\mathfrak{I}\supseteq\mathfrak{B}}\mathfrak{I}

and \mathfrak{T} the topology generated by \mathfrak{B}. Clearly since \mathfrak{T} is a topology on X which contains \mathfrak{B} we have that \mathfrak{T}\supseteq\mathfrak{J}. Conversely, let U\in\mathfrak{T}, then \displaystyle U=\bigcup_{\beta\in\mathcal{B}}S_\beta where \left\{S_\beta\right\}_{\beta\in\mathcal{B}}\subseteq\mathfrak{B}. But, for any topology T containing \mathfrak{B} we must have that \displaystyle \bigcup_{\beta\in\mathcal{B}}S_\beta\in T (since it’s  closed under arbitrary unions). Thus, \displaystyle \bigcup_{\beta\in\mathcal{B}}S_\beta is contained in all topologies which are supersets of \mathfrak{B}, namely it is in their intersection \mathfrak{J}. It follows that \mathfrak{J}=\mathfrak{T} as desired.

Now, let \mathfrak{B} be a subbasis and

\displaystyle \mathfrak{J}=\bigcap_{\mathfrak{I}\in T(X):\mathfrak{I}\supseteq\mathfrak{B}}\mathfrak{I}

and \mathfrak{T} the topology generated by \mathfrak{B}.

Clearly \mathfrak{T} being a topology on X containing \mathfrak{B} guarantees that \mathfrak{T}\supseteq\mathfrak{J}. Conversely, if U\notin\mathfrak{J}, then U\notin T for some topology T on X which contains \mathfrak{B}. But, this then means that T cannot be an arbitrary union of finite intersections of elements of \mathfrak{B}, since any topology T containing \mathfrak{B} must contain them. Thus, U\notin\mathfrak{T}. It follows that \mathfrak{T}=\mathfrak{J} as required.

6.

Problem: Show that the topologies \mathbb{R}_\ell and \mathbb{R}_K aren’t comparable.

Proof: We claim that (-1,1)-K\in\mathbb{R}_K but (-1,1)-K\notin\mathbb{R}_\ell. The fact that (-1,1)-K\in\mathbb{R}_K follows from the definition of the basis. Now, suppose there was a basis element [\alpha,\beta)\in\mathbb{R}_\ell such that 0\in[\alpha,\beta)\subseteq (-1,1)-K. Clearly \beta>0 otherwise 0\notin[\alpha,\beta). But, by the Archimedean principle there exists some n\in\mathbb{N} such that \displaystyle \frac{1}{n}<\beta, thus \displaystyle \frac{1}{n}\in[\alpha,\beta) but this contradicts that [\alpha,\beta)\subseteq (-1,1)-K. It follows that no such [\alpha,\beta) exists.

We now claim that [0,1)\in\mathbb{R}_\ell but [0,1)\notin\mathbb{R}_K. To see this suppose that (\alpha,\beta)-K is a basic element for \mathbb{R}_K such that 0\in (\alpha,\beta)-K\subseteq[0,1). Now, clearly \alpha<0 otherwise 0\notin(a,b)-K. But, by the Archimedean principle we may choose \displaystyle n\in\mathbb{N} such that \displaystyle \alpha<\frac{-1}{n}<0 and so \displaystyle \frac{-1}{n}\in(\alpha,\beta) and thus \displaystyle \frac{-1}{n}\in(\alpha,\beta)-K, but this contradicts that (\alpha,\beta)-K\subseteq[0,1)

7.

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

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

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

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

\mathfrak{J}_4=\text{ the upper limit topology, the topology with sets of the form }(\alpha,\beta]\text{ as a basis}

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

Determine, for each of these topology, which of the others contain it.

Proof:We know from the book that \mathfrak{J}_1\subseteq\mathfrak{J}_2.

Next, we claim that \mathfrak{J}_1\supseteq\mathfrak{J}_3. To see this let U\in\mathfrak{J}_3, then X-U=\{x_1,\cdots,x_n\} where we may assume WLOG that x_1<\cdots<x_n. So then, U=(-\infty,x_1)\cup(x_1,x_2)\cup\cdots\cup(x_n,\infty) which is the finite union of open sets in \mathbb{R}, thus U\in\mathfrak{J}_1

We claim that \mathfrak{J}_1\subseteq\mathfrak{J}_4. To see this we must merely note that

\displaystyle (\alpha,\beta)=\bigcup_{\gamma<\beta}(\alpha,\gamma]

so that every basic set, and thus every set, in \mathfrak{J}_1 may be written as the union of basic open sets in \mathfrak{J}_4 from where the conclusion follows.

It’s also clear that \mathfrak{J}_1\supseteq\mathfrak{J}_5 since

(-\infty,\alpha)=\bigcup_{\beta<\alpha}(\beta,\alpha)

and so every element of \mathfrak{J}_5 is the union of open sets in \mathfrak{J}_1 and so \mathfrak{J}_1\supseteq\mathfrak{J}_5

Now, clearly \mathfrak{J}_3\supseteq\mathfrak{J}_2\supseteq\mathfrak{J_1} and \mathfrak{J}_3\supseteq\mathfrak{J}_1\supseteq\mathfrak{J}_5 and so it suffices to check how it relates to \mathfrak{J}_4. We claim that they aren’t comparable. To see this we note that (0,1] is open in the upper limit topology, but any set of the form (a,b)-K must have b>1 and so it will entrap points not in (0,1] that aren’t removed by K. Also, using a very similar argument one can show that (-1,1)-K is not open in the upper limit topology.

The last inclusion we must check is how \mathfrak{J}_3 relates to \mathfrak{J}_5 and we claim they aren’t comparable. Clearly (-\infty,0)\notin\mathfrak{J}_2 sine it’s complement is infinite but it’s in \mathfrak{J}_5. Also, (-\infty,0)\cup(0,\infty)\in\mathfrak{J}_3 but it’s not in \mathfrak{J}_5 since any basic set containing 1, must contain 0

8.

a) Apply lemma 13.2 to show that the countable collection \mathfrak{B}=\left\{(a,b):a<b,\text{ }a,b\in\mathbb{Q}\right\}  is a basis that generates the standard topology on \mathbb{R}.

b) Show that the collection \mathfrak{C}=\left\{[a,b):a<b,\text{ }a,b\in\mathbb{Q}\right\} generates a topology different than \mathbb{R}_\ell

Proof:

a) Clearly since \mathfrak{B} is contained is the usual basis for \mathbb{R} the topology generated by it is coarser than the usual one. Conversely, let (a,b) be any basic open set in the usual topology on \mathbb{R} and x\in(a,b) arbitrary. We know from the structure of the real numbers that there is some a'\in\mathbb{Q} such that a<a'<x and some b\in\mathbb{Q} such that x<b<b'. Thus, (a',b')\in\mathfrak{B} and x\in(a',b')\subseteq(a,b) from where it follows that the topology generated by \mathfrak{B} is finer than the usual topology. Thus, it follows that the usual topology and the one generated by \mathfrak{B} are equal.

b) We claim that [\sqrt{2},3) is not in the topology generated by \mathfrak{C}. To see this suppose that [\sqrt{2},3) could be written as the union of sets of the form [\alpha,\beta),\text{ }\alpha,\beta\in\mathbb{Q}, say \displaystyle \bigcup_{\gamma\in\mathcal{C}}[\alpha_\gamma,\beta_\gamma). We see that since [\alpha_\gamma,\beta_\gamma)\subseteq[\sqrt{2},3) that \alpha_\gamma\geqslant\sqrt{2} and since it’s rational we may conclude that \alpha_\gamma>\sqrt{2}. But, this means that for every \gamma\in\mathcal{C} we have that \sqrt{2}\notin[\alpha_\gamma,\beta_\gamma) and thus \displaystyle \sqrt{2}\notin\bigcup_{\gamma\in\mathcal{C}}[\alpha_\gamma,\beta_\gamma) contradictory to our assumption.

Advertisements

September 30, 2010 - Posted by | Fun Problems, Munkres, Topology | , , , , , , ,

No comments yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: