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Recall that a compact Hausdorff space is second countable if and only if the Banach space $C(X)$ of continuous functions on $X$ is separable. I'm looking for a similar criterion for locally compact Hausdorff spaces, using $C_0(X)$ (the Banach space of continuous functions vanishing at infinity) instead of $C(X)$.

Naive Guess: Suppose $X$ is a second countable locally compact Hausdorff space. Then $C_0(X)$ is separable.

Proposed Proof: Since $X$ is second countable, so is its one-point compactification $\tilde{X}$. By the theorem above $C(\tilde{X})$ is is separable, and thus $C_0(X)$ is separable since subspaces of separable metric spaces are separable (not true for subspaces of arbitrary separable spaces, but OK for separable metric spaces). QED.

Problem: I'm not sure if the first line is true. This seems to be equivalent to the statement that every second countable locally compact Hausdorff space is the countable union of compact subspaces, and I can't find a proof or a counter-example. If this turns out to be false, what IS the right criterion for $C_0(X)$ to be separable?

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    $\begingroup$ The "Naive Guess" is true. See e.g. Kechris, Classical descriptive set theory, Theorem 5.3, page 29 for a number of equivalent reformulations of second countability of a locally compact space (among which $\sigma$-compact + metrizable). $\endgroup$ – t.b. Aug 20 '12 at 20:35
  • $\begingroup$ In the above argument, can we conclude that $C(X)$ is also separable metric space? Since $C(X) \subseteq C(\tilde{X})$ and subspaces of separable metric spaces are separable. $\endgroup$ – Pedro do Norte Apr 6 at 20:14
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Your argument is correct, thanks to the following result.

Theorem. For a locally compact Hausdorff space $X$ the following are equivalent:

  1. $X$ is Lindelöf.
  2. $X$ is $\sigma$-compact.
  3. There is a sequence $\langle K_n:n\in\Bbb N\rangle$ of compact subspaces of $X$ such that $K_n\subseteq\operatorname{int}K_{n+1}$ for each $n\in\Bbb N$ and $X=\bigcup_{n\in\Bbb N}K_n$.
  4. The point at infinity in the one-point compactification of $X$ has a countable local base.

Proof. Suppose that $X$ is Lindelöf. For each $x\in X$ let $U_x$ be an open nbhd of $x$ with compact closure, and let $\mathscr{U}=\{U_x:x\in X\}$. $\mathscr{U}$ is an open cover of $X$, so it has a countable subcover $\{V_n:n\in\Bbb N\}$. Clearly $X=\bigcup_{n\in\Bbb N}\operatorname{cl}V_n$ is a representation of $X$ as a countable union of compact sets, so $X$ is $\sigma$-compact.

Now assume that $X$ is $\sigma$-compact, and let $\{C_n:n\in\Bbb N\}$ be a countable cover of $X$ by compact sets. Let $K_0=C_0$. If the compact set $K_n$ has already been defined, for each $x\in K_n$ let $U_x$ be an open nbhd of $x$ with compact closure, and let $\mathscr{U}=\{U_x:x\in K_n\}$. $K_n$ is compact, so $\mathscr{U}$ has a finite subcover $\{V_1,\dots,V_m\}$. Let $K_{n+1}=C_n\cup\bigcup_{k=1}^m\operatorname{cl}V_k$; then $K_{n+1}$ is compact, and $K_n\subseteq\operatorname{int}K_{n+1}$. Finally, $C_n\subseteq K_n$ for $n\in\Bbb N$, so $\bigcup_{n\in\Bbb N}K_n=X$.

Now assume that there is a sequence $\langle K_n:n\in\Bbb N\rangle$ as in (3). Let $X^*$ be the one-point compactification of $X$, and let $p$ be the point at infinity. For $n\in\Bbb N$ let $B_n=X^*\setminus K_n$, and let $\mathscr{B}=\{B_n:n\in\Bbb N\}$; then $\mathscr{B}$ is a countable local base at $p$. To see this, let $U$ be any open nbhd of $p$ in $X^*$. Then $C=X^*\setminus U$ is a compact subset of $X$. $\bigcup_{n\in\Bbb N}\operatorname{int}K_n=X$, so $\{\operatorname{int}K_n:n\in\Bbb N\}$ is an increasing open cover of $C$, and therefore there is an $n\in\Bbb N$ such that $C\subseteq\operatorname{int}K_n\subseteq K_n$, whence $B_n\subseteq U$.

Finally, suppose that $X^*$ is first countable at $p$, the point at infinity, and let $\mathscr{B}=\{B_n:n\in\Bbb N\}$ be a nested local base at $p$. For $n\in\Bbb N$ let $K_n=X^*\setminus B_n$; then each $K_n$ is compact, and $X=\bigcup_{n\in\Bbb N}K_n$. Let $\mathscr{U}$ be an open cover of $X$. For each $n\in\Bbb N$ let $\mathscr{U}_n$ be a finite subset of $\mathscr{U}$ covering the compact set $K_n$, and let $\mathscr{V}=\bigcup_{n\in\Bbb N}\mathscr{U}_n$; then $\mathscr{V}$ is a countable subcover of $\mathscr{U}$. $\dashv$

Of course a second countable space is Lindelöf, so you get all of (1)-(4).

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  • $\begingroup$ Polish would be a useful thing to add to the list. An argument I would find nice would be to use Urysohn's lemma to embed $X$ into the Hilbert cube, then finish off by using that a locally compact and dense subset of a metric space is open (this also gives you $\sigma$-compactness of $X$ for free). $\endgroup$ – t.b. Aug 20 '12 at 20:59
  • $\begingroup$ @t.b.: What about $\omega\times(\omega_1+1)$? It’s locally compact and Lindelöf but not Polish. $\endgroup$ – Brian M. Scott Aug 20 '12 at 21:30
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    $\begingroup$ You're right of course, sorry about that (or any non-metrizable compact space, for that matter). I mixed up what you wrote with the equivalences given in Kechris's book: 1) metrizability of the 1-point compactification, 2) second countability, 3) Polish, 4) metrizable $\sigma$-compact and metrizable 5) homeomorphic to an open subset of a metrizable space, together with what I intended to write as an answer when you posted yours. $\endgroup$ – t.b. Aug 20 '12 at 21:36
  • $\begingroup$ That settles it! Thank you for your very detailed answer. $\endgroup$ – Paul Siegel Aug 21 '12 at 13:28

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