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Having some trouble with this question.

Let $X$ be a metric space in which every infinite subset has a limit point. Prove that $X$ is compact. Hint: By Exercises 23 and 24, $X$ has a countable base. It follows that every open cover of $X$ has a countable subcover ${G_n}$, $n = 1, 2, 3, ....$ If no finite subcollection of ${G_n}$ covers $X$, then complement $F_n$ of $G_1 \cup \dots \cup G_n$ is nonempty for each $n$, but $\bigcap F_n$ is empty. If $E$ is a set which contains a point from each $F_n$, consider a limit point of $E$, and obtain a contradiction.

I cannot justify the phrase "It follows that every open cover of $X$ has a countable subcover ${G_n}$, $n = 1, 2, 3, ....$", am I overlooking something simple? Is there a simple map between the countable base and any open subcover?

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    $\begingroup$ Let $\mathcal{B} = \{B_n\}$ be a countable base. For an open cover $\mathfrak{C} = \{C_\alpha \colon \alpha \in A\}$, consider $\mathcal{S} = \{ k \colon (\exists \alpha \in A)(B_k \subset C_\alpha)\}$. For each $k \in \mathcal{S}$, choose an $\alpha_k$ such that $B_k \subset C_{\alpha_k}$. The $C_{\alpha_k}$ form a countable subcover. (Proofs elementary but tedious.) $\endgroup$ – Daniel Fischer Jul 17 '13 at 0:06
  • $\begingroup$ I have another question for this exercise. Does anyone know why $\bigcap F_n$ is empty? $\endgroup$ – Hank Sep 30 '17 at 20:39
  • $\begingroup$ @DanielFischer Sorry for my ignorance, but I really want to know to prove it. $\endgroup$ – Hongyan Jun 2 '18 at 14:22
  • $\begingroup$ @Hongyan $\mathcal{S}$ is countable (as a subset of $\mathbb{N}$), hence $\{ C_{\alpha_k} : k \in \mathcal{S}\}$ is countable. And $\{ B_k : k \in \mathcal{S}\}$ is a cover of $X$, so $\{ C_{\alpha_k} : k \in \mathcal{S}\}$ a fortiori is a cover. To see that $\{ B_k : k \in \mathcal{S}\}$ covers $X$, let $x\in X$ arbitrary. Since $\mathfrak{C}$ is a cover, there is an $\alpha \in A$ with $x \in C_{\alpha}$. Since $C_{\alpha}$ is open and $\mathcal{B}$ is a base, there is a $k$ such that $x \in B_k \subset C_{\alpha}$. But then $k \in \mathcal{S}$. $\endgroup$ – Daniel Fischer Jun 6 '18 at 13:45
  • $\begingroup$ @Hank $F_n$ is defined as the complement of a finite subset of an open cover of $X$. Thus, the whole cover covers $X$. As you consider larger and larger subsets of the cover of $X$, the complement gets smaller and smaller. Eventually, if you take the union of all subsets $G_i$ of the open cover, you get the open cover. The complement of this is $\cap F_n$, which must be empty, since, by definition, we just covered $X$ $\endgroup$ – PhysMath Sep 23 '18 at 9:39
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Let $\mathscr{B}$ be a countable base for $X$, and let $\mathscr{U}$ be an open cover of $X$. For each $x\in X$ there is a $U_x\in\mathscr{U}$ such that $x\in U_x$, and there is a $B_x\in\mathscr{B}$ such that $x\in B_x\subseteq U_x$. Let $\mathscr{B}_0=\{B_x:x\in X\}$; clearly $\mathscr{B}_0$ is a countable cover of $X$. The construction of $\mathscr{B}_0$ ensures that for each $B\in\mathscr{B}_0$ there is a $U_B\in\mathscr{U}$ such that $B\subseteq U_B$; let $\mathscr{V}=\{U_B:B\in\mathscr{B}_0\}$. I claim that $\mathscr{V}$ is a countable subcover of $\mathscr{U}$.

  • Clearly $\mathscr{V}\subseteq\mathscr{U}$.
  • $\mathscr{V}$ is indexed by the countable set $\mathscr{B}_0$, so $\mathscr{V}$ is countable.
  • Let $x\in X$; by construction $x\in B_x\in\mathscr{B}_0$, so $x\in B_x\subseteq U_{B_x}\in\mathscr{V}$, so $\mathscr{V}$ covers $X$.
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    $\begingroup$ +1, I also had trouble with this part of the problem, and your answer was extremely helpful. My original strategy was to write each $U \in \mathscr{U}$ as a union of $B \in \mathscr{B}$, and then show that the collection of all possible unions of $B \in \mathscr{B}$ is countable. Oops--this collection is actually uncountable. $\endgroup$ – Elliott May 17 '14 at 20:21
  • $\begingroup$ Do you know why $\bigcap F_n$ is empty? $\endgroup$ – Hank Sep 30 '17 at 20:40
  • $\begingroup$ Sorry. I don't know why "for each $B\in\mathscr{B}_0$ there is a $U_B\in\mathscr{U}$ such that $B\subseteq U_B$". $\mathscr{B}=\mathscr{B}_0$, right? I know for each $U_x$,there is a $B_x$ such that $B_x\subseteq U_x$. But I am not sure the converse is true. $\endgroup$ – Hongyan Jun 4 '18 at 13:16
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I do not know if this is what Rudin had in mind, but if we let $\mathcal{Y}$ be an open cover of $X$. Then since $X$ has a countable base $\mathcal{B}$, each open set in $\mathcal{Y}$ can be written as a union of (at most) countably many elements in $\mathcal{B}$. So you can choose a countable subcover of $\mathcal{Y}$ by choosing open sets of the form $\mathcal{Y}_{\alpha}:B_{\alpha}\in \mathcal{Y}_{\alpha}$. Here $\alpha$ is an index set such that $\bigcup B_{\alpha}=X$. The choice of $\mathcal{\alpha}$ may not be unique and some $\mathcal{Y}_{\alpha}$ might even coincide or empty, but after adjusting indexes we can get a countable subcover as desired. This construction seems identical with the comment above, though.

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  • $\begingroup$ I was with you until you said "So you can choose a countable subcover...", I am having a hard time justifying why you can just "choose" the right ones to make everything work, maybe its just my imagination complicating things though... $\endgroup$ – user86589 Jul 17 '13 at 3:57
  • $\begingroup$ I suspect you did found a typo in my post. The original answer would not make sense as $B_{\alpha}\in \mathcal{Y}$ is automatic. $\endgroup$ – Bombyx mori Jul 17 '13 at 4:01
  • $\begingroup$ I'm assuming $B_{\alpha}$ are from the countable base, is that correct? In that case, what is to say that some of these are a subset of any (finite number of) subset of $Y$? $\endgroup$ – user86589 Jul 17 '13 at 4:10
  • $\begingroup$ I mean $B_{\alpha}$ is contained in $Y_{\alpha}$. There may be multiple $Y_{\alpha}$s containing $B_{\alpha}$, so it suffice to just pick up one. $\endgroup$ – Bombyx mori Jul 17 '13 at 4:12
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Your question, in clear words and proper terminology, is how to show that every second-countable topological space (or even just metric space) is Lindelöf.

To see that note, that if $\cal U$ is an open cover and $\mathcal B=\{V_n\mid n\in\Bbb N\}$ is a countable basis, then taking $\mathcal V_n=\{U\in\mathcal U\mid V_n\subseteq U\}$ is a countable family. It might be that some $V_n$ are empty, in which case we omit them. But we are left with a family of at most $\aleph_0$ non-empty sets. Pick $U_n\in\mathcal V_n$.

I claim that $\{U_n\mid n\in\Bbb N\}$ is an open cover. Those are all open sets, so we just need to show it's a cover. Let $x\in X$ be an arbitrary point, then for some $n\in\Bbb N$ we have that $x\in V_n$, and for some $U\in\mathcal U$ we have that $x\in U$. Therefore $U\cap V_n$ is a non-empty open set and so there is some $m$ such that $x\in V_m\subseteq U\cap V_n$.

Therefore $x\in U_m$ by the definition of $\mathcal V_m$, as wanted.

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