Show A is locally compact, if A is a closed subset of a locally compact space. Given X a locally compact space and $A \subset X$ a closed subset. How do I show that A is locally compact as well?
I have no clue whatsoever on how I should handle this!
 A: Let $x\in A$, a there exists a compact neighborhood $C\subset X$ of $X$ of $x$ in $X$. $C\cap A$ is a compact neighborhood of $x$ in $A$. We deduce that $A$ is locally compact.
To show that $C\cap A$ is compact, consider a covering $(U_i)_{i\in I}$ of $C\cap A$. You can suppose that there exists an open subset $V_i$ of $X$ such that $U_i=V_i\cap A$, Since $A$ is closed, $A\cap C$ is closed and $W=X-A\cap C$ is open. This implies that $W\bigcup \bigcup_{i\in I}V_i$ is an open covering of $C$. You can extract a finite covering $W, V_1,...,V_n$. This implies that $U_1,..U_n$ cover $A\cap C$.
A: Here is a result that might help you in answering problems like these (well, I find it useful). 
Let $(X,\tau)$ be a Hausdorff space and $\emptyset\neq S\subseteq X$. Then $(S,\tau_S)$ is locally compact if and only if for each $x\in S,$ there exists a nbhd $V$ of $x$ in $X$ such that $\overline{V\cap S}\subseteq S$ and $\overline{V\cap S}$ is compact.
Proof:($\Rightarrow$) Suppose $(S,\tau_S)$ is locally compact. Let $x\in S$. Then there exists a nbhd $U$ of $x$ in $S$ such that $\overline{U}^S$ is compact. Since $\overline{U}^S$ is compact it is closed, whence $\overline{U}^S=\overline{U}$. Since $U$ is open in $S$ there exists $V\in\tau$ such that $U=S\cap V$, and $V$ is a nbhd of $x$ in $X$. Now $\overline{S\cap V}\subseteq \overline{U}=\overline{U}^S\subseteq S$ and since $\overline{U}^S$ is compact $\overline{S\cap V}$ is compact.
$(\Leftarrow)$ Conversely suppose the given condition. Let $x\in S$. Then $V\cap S$ is a nbhd of $x$ in $S$ such that $\overline{V\cap S}^S$ is compact. 
Now you can use the above result to solve your problem.
Let $x\in A$. Since $X$ is locally compact there exists a nbhd $U$ of $x$ in $X$ such that $\overline{U}$ is compact. Then $\overline{U\cap A}\subseteq\overline{A}=A$ as $A$ is closed, and $\overline{U\cap A}$ is compact as it is a closed subset of $\overline{U}$ which is compact. Now by the above result $A$ is locally compact. 
You can in fact prove that $A$ is locally compact if $A$ is


*

*Open,

*Intersection of a closed and an open set.

