intersection of two relatively compact spaces It is known that intersection of two compact spaces is might not compact but intersection of two compact Hausdorff spaces is compact.
I curious about intersection of two relatively compact spaces.
In metric spaces, compact and relatively compact are equivalent. So, there is no issue in metric spaces.
My question is: is intersection of two relatively compact spaces relatively compact? If not, please show me an counterexample.
 A: Note that the intersection of two compact Hausdorff spaces need not be compact:

Let $X=((0,1),\tau)$, the open interval with the Euclidean topology. Let $A=X\cup\{a\}$ and $B=X\cup\{b\}$, where $a$ and $b$ are distinct points outside of $X$. On $A$, let the topology $\tau_A$ consist of the open sets in $X$, together with sets of the form $A\setminus K$, where $K$ is a compact subset of $X$. The topology $\tau_B$ on $B$ is built in a similar way. Then both $(A,\tau_A)$ and $(B,\tau_B)$ are homeomorphic to the circle $S^1$, thus are compact Hausdorff spaces. We can form the space $A\cup B$ by taking as open sets those sets $U$ for which $U\cap A$ and $U\cap B$ is open. Then $A\cup B$ is a compact space but not Hausdorff, and $A\cap B\approx X$ is not compact.

But I guess that you actually meant that the intersection of two compact sets in a Hausdorff space is compact. That is true.
It is also true that the intersection of two relatively compact sets is relatively compact. In fact, any subset of a relatively compact space is relatively compact, for if $B\subset A$ and $\overline A$ is compact, then $\overline B\subseteq\overline A$ is a closed subset of a compact space, thus compact.
