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?

  • 1
    $\begingroup$ 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

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 $X$.
  • 1
    $\begingroup$ +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

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.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.