A $T_2$ space is locally compact iff it has a base $\beta $ s.t $\forall B\in \beta $ we have $\bar B$ is compact. A $T_2$ space is locally compact iff it has a base $\beta $ s.t $\forall B\in \beta $ we have
$\bar B$ is compact.
My definition of locally compact is that $\forall x\in X$ has a compact neighborhood in $X$.
$(\Rightarrow)$ So let $ U$ be an open set in $X$, we show that $U$ is equal to the union of basic open sets $B$ from $\beta$, $U=\cup B, B\in \beta$ and $\bar B$ is compact.
Let $x \in U$ and let $U'$ be the compact nbhd of $x$, since $U$ is open then $U$ is not the singleton set since we are in $T_2$, therefore there is at least another element $y$ in $U$ and $x\neq  y$, and since we are in $T_2$ there exist a open nbhd of $y$, say $V$ that is disjoint from $U$, and let $V'$ be the compact nbhd of $y$..Now I'm lost
 A: $\newcommand{\cl}{\operatorname{cl}}$The start of your argument doesn’t match the direction that you say that you’re trying to prove. For the implication $(\Rightarrow)$ you don’t have a base of open sets with compact closure: you have to construct one on the assumption that $X$ is locally compact and $T_2$.
HINT: Suppose that $\langle X,\tau\rangle$ is a locally compact $T_2$-space. Then each $x\in X$ has a compact nbhd $N_x$. Let $U_x=\operatorname{int}N_x$; then $U_x$ is an open nbhd of $x$, and $\cl U_x\subseteq N_x$, so $\cl U_x$ is compact and therefore $T_3$ (why?). Use this to show that $X$ itself is $T_3$. Now let $$\mathscr{B}=\{V\in\tau:\cl V\text{ is compact}\}\;,$$ and show that $\mathscr{B}$ is a base for $\tau$. That is, show that if $x\in U\in\tau$, then there is a $B\in\mathscr{B}$ such that $x\in B\subseteq U$.
The other direction is trivial: for it you assume that the space has a base $\mathscr{B}$ of open sets with compact closure and show that each point of $X$ has a compact nbhd.
A: HINT： $x$ has a compact nbhd $U$, where $\overline{U}$ is compact; then $\exists B \in \mathcal B$, such that $x \in B \subseteq U$. So $\overline{B}$ is also compact, since $\overline{B} \subseteq \overline{U}$.
