There are several definitions of local compactness - from

  1. "every point has a compact neighbourhood",
  2. "every point has a base of compact neighbourhoods" to Hatcher's:
  3. "every neighbourhood contains a compact neighbourhood".

It is easy to see that the second implies the third and vice versa. I can't see why the first (+Hausdorff) is equivalent to them. I would like to know why compactness (assuming Hausdorff) implies local compactness (defined by 3.) and implication 1=>2&3 seems useful here ;)

  • $\begingroup$ I remember this being a not so obvious proof; the one I know rrlies on the trivial fact that in a Hausdorff space, the intersection of the closures of all neighborhoods of a point is that point. I might write down the proof later. $\endgroup$ – Olivier Bégassat Dec 14 '12 at 17:31
  • $\begingroup$ Or you can look up the proof in the chapter on local compactness in Frédéric Paulin's excellent lecture notes available online called "Topologie et Calcul différentiel"; although in french, you don't need to speak it to understand his proof. $\endgroup$ – Olivier Bégassat Dec 14 '12 at 17:34
  • 1
    $\begingroup$ You can do it in two steps: The first one (the harder step) is to show that a Hausdorff space with a compact neighborhood around each point is regular. The second step (which is very easy) is to show that every regular space with property 1 also satisfies 3. During the first step you'll need the fact that a compact Hausdorff space is regular. $\endgroup$ – Stefan Hamcke Dec 14 '12 at 17:40
  • $\begingroup$ @OlivierBégassat Do you mean these notes: math.u-psud.fr/~paulin/notescours/cours_analyseI.pdf ? It's hard for me to find this theorem, which one is it? $\endgroup$ – savick01 Dec 14 '12 at 19:35
  • $\begingroup$ Btw. I'll try your approach @StefanH. Thanks for help. $\endgroup$ – savick01 Dec 14 '12 at 19:54

We'll follow the sketch by Stefan H. (see comments) - we'll show that a Hausdorff space $X$ satisfying 1. satisfies 3.

First - let's show that a Hausdorff space $X$ satisfying 1. is regular i.e. any closed set $C$ and an external point $p$ can be separated by neighbourhoods.

Let's consider a compact neighbourhood $U$ of $p$. It is a well known fact that a closed set and a point (even another disjoint closed set) can be separated by neighbourhoods in a compact Hausdorff space, so there are disjoint neighbourhoods (in $U$) $D$ of $C\cap U$ and $V$ of $p$.

We can choose $V\cap \mathrm{int}U$ as a neighbourhood of $p$ in $X$. A disjoint neighbourhood of $C$ is $X\setminus U \cup D$. It is open: $X\setminus U$ is open, so we only need to notice that any point $d$ of $D$ has a neighbourhood contained in $X\setminus U \cup D$ - $d$ has a neighbourhood contained in $D$ open in $U$, which is an intersection of some neighbourhood $E$ open in $X$ - that neighbourhood is obviously contained in $X\setminus U \cup D$.

Now a small lemma (we don't need the "if" part actually):

A space $X$ is regular iff every point has base of closed neighbourhoods.

$\Rightarrow$ It suffices to prove that every open neighbourhood $U$ of a point $p\in X$ contains a closed one. $X\setminus U$ is closed so it can be separated from $p$ by open sets $V,W$ respectively. $X\setminus V\supseteq W$ is a closed neighbourhood of $p$ contained in $U$.

$\Leftarrow$ Consider a closed set $C$ and $p\notin C$. Since $X\setminus C$ is open, $p$ has a closed neighbourhood $P$ disjoint with $C$. The separating open sets are: $X\setminus P$ and $\mathrm{int} P$.

All we need now is to notice that since any neighbourhood $U$ of $p$ contains a closed neighbourhood $V$ and the intersection of $V$ with a compact neighbourhood W of $p$ (given by assumption 1.) is compact, then $U$ contains a compact neighbourhood $V\cap W$.

  • $\begingroup$ Everything's correct ;-) $\endgroup$ – Stefan Hamcke Dec 15 '12 at 12:19

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