$\forall x\in X,\exists(x\in)U\subset X$ s.t. $\overline{U}$ is compact, $\iff$ $\forall x\in X,U\subset X,\exists V$ s.t. $\overline{V}$ is compact Let $X$ be a Hausdorff space. Show that the following statements are equivalent: 


*

*$\forall x\in X,\exists(x\in)U\subset X$ such that $\overline{U}$ is compact.

*$\forall x\in X,x\in U\subset X,\exists V$ such that $\overline{V}\subset U$   is compact.


I am quite clueless here. I know that if, given $x\in X$ and $x\in U\subset X$, if the first statement holds, then there has to be $V$ neighborhood of $x$ such that $\overline{V}$ is compact. Whether $\overline{V}\subset U$, $U\subset \overline V$, or $V\setminus U\ne \emptyset$, I only have $\overline{V}\cap U$ to look at. as the required set. But I still feel like I am merely making a non-profound guess, and I can't quite see the relevance of $X$ being Hausdorff to the process I attempted. Could you expand my horizons in that topic?  
 A: Hint: In Hausdorff spaces the following holds: Let $K_1, K_2$ be compact sets, then there exist open sets $U_1, U_2$ such that $K_1\subseteq U_1, K_2\subseteq U_2$ and $U_1\cap U_2=\emptyset$.
Apply this to the sets $\{x\}, \partial U$ to show that the first statement implies the second.
Remark: The Hausdorff property is crucial as it gives you the nice separation above. The following example might be instructive: Endow ${x, y} with the topology
$$ \{  \{x\}, \{x,y\} \}.$$
Can you find a $U$ such that the second statement fails?
A: Now first I will claim the following is true for $X$

For a compact set $K$ and a point $x\notin K$ there exists neighborhood $U_x$ and $V_K$ such that $U_x\cap V_K=\emptyset$

Since $X$ is Hausdorff, for each $y\in K$ we can find two disjoint neighborhoods $B_y$ and $B_x(y)$ such that $B_x\cap B_y=\emptyset$. $\{B_y\}$ is open covering of $K$ thus $\{B_{y_1},\dots B_{y_n}\}$ is a finite subcover for $K$. Set $U_x=\cap_{i} B_x (y_i)$ and $V_K=\cup_i B_{y_i}$. It is easy to show that $U_x\cap V_K=\emptyset$
Now suppose $x\in X$ and $U_x\subset X$. By the first statement there exists $W_x$ such that $\bar{W}$ compact. Take $T=U\cap W$ and consider the set $\partial T$ of boundary points of $T$. This set is closed and contained in $\bar{T}\subset \bar{W}$. Since $X$ is Hausdorff this implies that $\partial T$ is compact.
Now  I claim that 

There exists $V_x$ such that $\bar{V}\subset \partial T$

Since $\partial T$ is compact by the first result we can find disjoint neighborhoods $N$ of $x$ and $M$ of $\partial T$.This implies $\bar{N}\cap \partial T=\emptyset$. Let $V=T\cap N$. Then
$$
\bar{V}\subset \bar{T}\cap \bar{N}=(T\cup \partial T)\cap \bar{N}=T\cap\bar{N}\subset T
$$
Then the claim follows.
A: 1.0. Let $U$ be a nbhd of $x.$ Let $U'$ be a nbhd of $x$  such that $$A=Cl_X( U')$$ is compact.Let $U''=U\cap U'$ and let $U'''$ be open in $X$ with $x\in U'''\subset U''.$
1.1.  $U'''$ is open in $X$ so it is open in the subspace $A$.
1.2.   Key point: Any compact Hausdorff space is a regular space. We have $x\in U'''$ where $U'''$ is open in $A.$ By regularity of $A$ there exists $V,$ open in $A,$ such that  $$x\in V\subset  Cl_A(V)\subset U'''.$$ 1.3. Since $V$ is open in $A$ and   $V\subset U'''\subset A$ with $U'''$ open in $X$,  we know that $V$ is open  in $X.$
1.4 .  $Cl_A(V)=Cl_X(V)$ (because $A$ is closed in $X$) and it is compact, because it  is closed in the compact  Hausdorff space $A$.
1.5. Putting all this together we have $$x\in V\subset Cl_X(V)\subset U,$$ where  $V$ is open in $X,$ and $Cl_X(V )$ is compact.
2.0. For the reverse implication, if $U$ is a nbhd of $x$ and there exists a nbhd $V$ of $X$ such that $\overline V$ is compact and  $x\in V\subset \overline V\subset U,$ then to find a nbhd $U'$ of $x$ such that $\overline {U'}$ is compact,let $U'=V.$ If you are confused here then change $U$ in your first displayed line to $U'$.
Remark. More strongly than the "key point" (1.2): A compact Hausdorff space is a  normal space.
