# Why must a locally compact second countable Hausdorff space be second countable to imply paracompactness?

The textbook version of the result I've seen states: A locally compact second countable Hausdorff space is paracompact. Is the property of being second countable needed, or have I missed something?

My thinking:

• If the space is locally compact then each point has a compact neighborhood.

• For this compact neighborhood each covering has a finite sub-covering.

• The finite sub-covering is a locally finite refinement.

• The last step is in error. For each point you get a finite subcover of a compact neighborhood of that point. But to get back to a cover of the entire space, you have to put all those subcovers back together, and the resulting cover need not be finite or even locally finite. If you have second countability, you can put them back together in a more careful way. – Nate Eldredge Feb 3 '15 at 23:28
• You certainly can't hope to find a finite refinement in general, even with second countability. Consider the covering of $\mathbb{Z}$, with its discrete topology, by singletons. – Nate Eldredge Feb 3 '15 at 23:30
• Edited the question, it was supposed to read locally finite refinement. – CuriousGeorge Feb 3 '15 at 23:47
• Thanks everyone! After doing a bit of reading based on the comments this all makes a lot more sense. – CuriousGeorge Feb 6 '15 at 23:42

Overall the hypotheses are gross overkill, but you can’t omit any of them. Matt Samuel has already given the example of the long line to show that you can’t omit second countability.

The particular point topology on a countably infinite set $X$ is non-Hausdorff, locally compact, and second countable — if $p$ is the particular point, $\mathscr{B}=\big\{\{p,x\}:x\in X\big\}$ is a countable base for the topology — and it’s not paracompact, since the open cover $\mathscr{B}$ has no open refinement that is locally finite at $p$. This shows that you can’t omit Hausdorffness.

To see that you can’t omit local compactness, let

\begin{align*} D&=\left\{\left\langle\frac1m,\frac1n\right\rangle:m,n\in\Bbb Z^+\right\}\;,\\\\ H&=\left\{\left\langle\frac1m,0\right\rangle:m\in\Bbb Z^+\right\}\;,\text{ and}\\\\ X&=\{\langle 0,0\rangle\}\cup H\cup D\;. \end{align*}

Points of $D$ are isolated. For $m,n\in\Bbb Z^+$ let $x_m=\left\langle\frac1m,0\right\rangle$ and $y_{m,n}=\left\langle\frac1m,\frac1n\right\rangle$, and define

$$B_n(x_m)=\{x_m\}\cup\{y_{m,k}:k\ge n\}\;;$$

$\{B_n(x_m):n\in\Bbb Z^+\}$ is a local base at $x_m$. (In other words, the sequence $\langle y_{m,n}:n\in\Bbb Z^+\rangle$ converges to $x_m$.) Let $p=\langle 0,0\rangle$, and for $n\in\Bbb Z^+$ let

$$B_m(p)=\{y_{k,n}:k\ge m\}\;;$$

$\{B_m(p):m\in\Bbb Z^+\}$ is a local base at $p$. With this topology $X$ is Hausdorff and second countable, but it’s not locally compact at $p$. It’s also not paracompact: the open cover

$$\mathscr{U}=\{B_1(p)\}\cup\{B_1(x_n):n\in\Bbb Z^+\}$$

has no open refinement that is locally finite at $p$.

The only real function served by local compactness is to make the space regular: every locally compact Hausdorff space is regular. If we require the space to be regular and Hausdorff, we can drop the assumption of local compactness, because every second countable space is Lindelöf, and every $T_3$ Lindelöf space is paracompact. Second countability is of course a good deal stronger than the Lindelöf property, so this theorem is really just a weak version of the theorem that a $T_3$ Lindelöf space is paracompact.

Without the second countable condition the result is not true. The long line is locally compact Hausdorff and is not paracompact because it is locally metrizable but not metrizable.

• It may be necessary for connected spaces, but it isn't for disconnected spaces, $\mathbb{R}\times D$ is paracompact for every discrete $D$. – Daniel Fischer Feb 3 '15 at 23:14
• @DanielFischer I don't know whether it's necessary to imply paracompactness, I meant to illustrate that the result isn't true without the second countable assumption. – Matt Samuel Feb 3 '15 at 23:16
• And it's a good example. But something should be done to the first sentence. I remember that a manifold (a locally Euclidean Hausdorff space) is paracompact if and only if every connected component is second countable. But I don't know how it is for general locally compact Hausdorff spaces. – Daniel Fischer Feb 3 '15 at 23:23
• @DanielFischer I've edited my answer. – Matt Samuel Feb 3 '15 at 23:25
• Alternatively, the long line contains a closed copy of $\omega_1$, which isn’t even metacompact, thanks to the pressing-down lemma. For that matter, you could just use $\omega_1$, since it’s locally compact and Hausdorff. – Brian M. Scott Feb 3 '15 at 23:53