Different notions of local compatness and why are they implied by compactness? There are several definitions of local compactness - from


*

*"every point has a compact neighbourhood",

*"every point has a base of compact neighbourhoods" to Hatcher's:

*"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 ;)
 A: 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 $V\cap W$. 
