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Let $X$ be a locally compact, second countable Hausdorff space. I want to prove that it has a countable basis of opens with compact closure, and that this basis can be extracted as a subset of any basis of its topology by restricting to opens with compact closure.

I understand the following:

As locally compact space, for every point $x$ of $X$ we can find a compact neighbourhood $V$, thus, for every point $x$ of $X$ we can find an open $V_x$ containing $x$ and whose closure is compact (by taking $V_x = V$ if $V$ is open, or the open set whose closure is $V$ if $V$ is closed, i.e: $V_x$ such that $V_x = V°$).

As second countable space, there is a basis family $\{U_{\alpha}\}_{\alpha \in A}$ of opens such that for every point $x$ of $X$, and every open $V_x$ containing $x$, there is a $\alpha\in A$ such that $x\in U_{\alpha}\subset V_x$.

Now, take a point $x$ in $X$, we can find an open $V_x$ around $x$ whose closure is compact, and find an element $U_{\alpha}$ of the basis such that $x\in U_{\alpha}\subset V_x$, the closure of $U_{\alpha}$ is a closed subset of the compact closure of $V_x$, thus, it is also compact.

Thus, for each point $x$ and for each open $V_x$ around that point whose closure is compact we can find an open $U$ with compact closure containing $x$ and contained in $V$.

Opens of the family basis whose closures are compact form a new basis family for the opens of the topological space with compact closures.

Now, how to show that it actually form a new basis family for all the opens of the topological space? I guess that the "Hausdorff" condition has something to do with.

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    $\begingroup$ Regarding the second paragraph: It is not always possible to find an open set whose closure is a given closed set. Closed sets that are closures of open sets are called regular closed sets. These are also equal to the closure of their own interior. But this does not cause a problem, just take $V_x=$int$V$. $\endgroup$ – Stefan Hamcke Jul 31 '13 at 13:43
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I want to prove that it has a countable basis of opens with compact closure,

That part is easy,

and that this basis can be extracted as a subset of any basis of its topology by restricting to opens with compact closure.

and this part is wrong as written, given the countable basis of the topology of $\mathbb{R}$ formed by the intervals with rational endpoints, we can easily construct an uncountable family of pairwise disjoint countable bases of the topology by shifting all intervals by a fixed irrational amount.

For the easy part, we formulate the

Lemma: Let $(X,\,\tau)$ be a locally compact Hausdorff space. For any basis $\mathcal{B}$ of $\tau$, the family $\mathcal{B}_c = \{ U \in \mathcal{B} : \overline{U} \text{ is compact}\}$ is also a basis of $\tau$.

Proof: Since $\mathcal{B}_c \subset \tau$, we only need to show

$$V \in \tau \Rightarrow V = \bigcup_{U \in \mathcal{B}_c,\, U \subset V} U,$$

or, that for every open $V$ and every $x \in V$, there is a $U \in \mathcal{B}_c$ with $x \in U \subset V$. Now, $\tau$ is locally compact, hence there is a compact neighbourhood $K_x$ of $x$ contained in $V$(1). Since $\mathcal{B}$ is a basis of $\tau$, there is a $U_x \in \mathcal{B}$ with $x \in U_x \subset \overset{\circ}{K}_x$. Since $\tau$ is Hausdorff, compact sets are closed, hence $\overline{U}_x \subset K_x$ is also compact, and therefore $U_x \in \mathcal{B}_c$. $\hphantom{foobarbazquuxbongbambooooopppppppppppppp}$ c.q.f.d.

If we start with a countable basis of the topology, extracting the relatively compact basis sets yields a countable basis of relatively compact sets.

Now the question remains whether in a second countable locally compact Hausdorff space every basis of the topology contains a countable basis.

That is the case for every second countable space (thanks to Stefan H. for the simple general proof).

Let $\mathcal{C}$ be a countable basis, and $\mathcal{B}$ any basis of the topology. Consider the set

$$\Gamma = \{ (C_1,\, C_2) \in \mathcal{C}^2 : (\exists B \in \mathcal{B})(C_1 \subset B \subset C_2)\}.$$

As a subset of the countable set $\mathcal{C}^2$, it is countable. For every $p = (C_1,\,C_2) \in \Gamma$, choose a $B_p \in \mathcal{B}$ with $C_1 \subset B_p \subset C_2$. Then $\mathcal{B}_\Gamma = \{ B_p : p \in \Gamma\}$ is a countable subset of $\mathcal{B}$. It is also a basis of the topology, since for open $V \neq \varnothing$ and $x \in V$, we can find a $C_2 \in \mathcal{C}$ with $x \in C_2 \subset V$, since $\mathcal{C}$ is a basis. Since $\mathcal{B}$ is a basis, there is a $B \in \mathcal{B}$ with $x\in B \subset C_2$. Since $\mathcal{C}$ is a basis, there is a $C_1 \in \mathcal{C}$ with $x \in C_1 \subset B$. Then $(C_1,\,C_2) \in \Gamma$, and hence

$$x \in C_1 \subset B_{(C_1,\,C_2)} \subset C_2 \subset V.$$

Thus, in a second countable locally compact Hausdorff space, every basis of the topology contains a countable basis of relatively compact sets.


(1) In a locally compact Hausdorff space $X$, every point has a neighbourhood basis of compact sets.

For $x \in X$, let $K_x$ be a compact neighbourhood of $x$.

Let $V$ be an arbitrary open neighbourhood of $x$. Then $U = V \cap \overset{\circ}{K}_x$ is an open neighbourhood whose closure is compact (since it is contained in $K_x$, which is closed because $X$ is Hausdorff). Thus $\partial U$ is a compact set, and $x \notin \partial U$. For every $y \in \partial U$, there are disjoint open neighbourhoods $W_y$ of $x$ and $Z_y$ of $y$. $\partial U$ is compact, hence there are $y_1,\, \ldots,\,y_k$ such that $\partial U \subset Z := \bigcup_{j=1}^k Z_{y_j}$. $W := \bigcap_{j=1}^k W_{y_j} \cap U$ is an open neighbourhood of $x$ with $W \cap Z = \varnothing$, and $W \subset U$. Hence $\overline{W} \subset \overline{U}\setminus Z \subset \overline{U}\setminus \partial U = U \subset V$.

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    $\begingroup$ It is indeed true that each base contains a base of minimal cardinality. Theorem: Let $\mathcal B$ be a base and $\mathcal C$ a countable base. Then $\mathcal B$ contains a countable base: Proof: Let $\Gamma$ be the set of pairs $(C_i,C_j)\in\mathcal C\times\mathcal C$ such that there is a set $B\in\mathcal B$ with $C_i\subseteq B\subseteq C_j$. For each pair $(C_i,C_j)\in\Gamma$ (there are countably many) choose a set $B_{ij}\in\mathcal B$ such that $C_i\subseteq B_{ij}\subseteq C_j$. Let $\mathcal D$ be the set of all these $B_{ij}$ ... $\endgroup$ – Stefan Hamcke Jul 31 '13 at 14:21
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    $\begingroup$ ... We will show that $\mathcal D$ is a base. For an open set $U$ and $x\in U$, there are $C_i, C_j\in\mathcal C,B\in\mathcal B$ such that $x\in C_i\subseteq B\subseteq C_j\subseteq U$. But this means that $(C_i,C_j)\in\Gamma$ and thus $x\in B_{ij}\subseteq U$. So $\mathcal D\subseteq\mathcal B$ is a base and has countable cardinality. $\endgroup$ – Stefan Hamcke Jul 31 '13 at 14:22
  • $\begingroup$ Ah, nice, thanks. $\endgroup$ – Daniel Fischer Jul 31 '13 at 14:24
  • $\begingroup$ Not yet nice for me :), I do not quite understand why for every open $V$ and every point $x\in V$ there is a compact neighbourhood of $x$ contained in $V$, to my knowledge, local compactness means that there is a least one compact neighbourhood for every point $x$, taking an arbitrary open $V$ around that point does not make it contained in that compact neighbourhood, nor containing it! $\endgroup$ – ubugnu Jul 31 '13 at 14:37
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    $\begingroup$ @ubugnu We're Hausdorff, so every $x$ has a neighbourhood basis of compact sets. Let $K_x$ be a compact neighbourhood of $x$. Let $V$ be any open neighbourhood of $x$. $U = V\cap \overset{\circ}{K}_x$ is an open neighbourhood of $x$ whose closure is compact (since it is contained in $K_x$). $\partial U$ is a compact set disjoint from $x$, so there are open sets $W, Z$ with $x \in W,\, \partial U \subset Z,\, W\cap Z = \varnothing$. Without loss of generality, $W \subset U$ (use $W\cap U$ otherwise). Then $\overline{W} \subset U \subset V$, and $\overline{W}$ is compact. $\endgroup$ – Daniel Fischer Jul 31 '13 at 14:47
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I don't usually think about such questions, but this appears to me to be a simplification of Daniel Fischer's proof of the first fact:

Let $\mathcal{B}$ be a countable basis. It suffices to show that for each open set $V$ and each $x \in V$ there is a relatively compact $W$ in $\mathcal{B}$ such that $x \in W \subset V$.

Now given such an open $V$ and $x \in V$, local compactness ensures that there is a compact $K$ such that $x \in K$ and $x \in K^{o}$. Consider $ U = V \cap K^{o}$. Since $U$ is open and $\mathcal{B}$ is a basis there exists some $W \in \mathcal{B}$ such that $x \in W \subset U$. Since $W \subset K^{o}$ we have that $\bar{W} \subset \bar{K^{o}}$, and since the space is Hausdorff and $K$ is compact $\bar{K^{o}} = K$. Thus $\bar{W}$ is a closed subset of a compact set, and therefore compact. Thus $W$ is relatively compact, completing the proof.

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  • $\begingroup$ I suspect this proof is incorrect, or at least uses incorrect reasoning. It is not true that in a Hausdorff space, every compact set $K$ satisfies $\overline{(K^\circ)} = K$. Consider the set $K = [0,1] \cup \{ 2 \}$. Then $K^\circ = (0,1)$ and $\overline{(K^\circ)} = [0,1]$. $\endgroup$ – Ben Bray Jan 20 at 17:44
  • $\begingroup$ I think this can actually be fixed by noting that since $W \subset \overline{(K^\circ)} \subset K$, $\overline{W}$ is still a subset of a compact set. So the reasoning goes through. $\endgroup$ – Ben Bray Jan 20 at 17:55

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