# Is the intersection of two locally compact subspaces locally compact?

Taking locally compact as such that every point has a local base of compact neighborhoods, is the intersection of two locally compact subspaces locally compact?

Let $$X=\bigl((\mathbb R\setminus\mathbb Q)\times\{-1,1\}\bigr)\cup\bigl( \mathbb Q\times\{0\}\bigr)$$ where we declare $U\subseteq X$ open iff $U=(V\times\{-1,0,1\})\cap X$ for some open subset $V$ of $\mathbb R$. Admittedly, $X$ is not Hausdorff. But it contains two subsets $(\mathbb R\times\{0,1\})\cap X$ and $(\mathbb R\times\{0,-1\})\cap X$ that are homeomorphic to the locally compact space $\mathbb R$, whereas their intersection $(\mathbb R\times\{0,\})\cap X$ is homomorphic to the not locally compact space $\mathbb Q$.
• You can make it $T_1$: it still works if you replace this topology by its join with the cofinite topology. – Brian M. Scott May 28 '15 at 12:41
The answer is positive for Hausdorff spaces. Let $X$ be a Hausdorff topological space, $A$, $B$ be locally compact subspaces of the space $X$, and $x\in A\cap B$ be an arbitrary point. Let $U\ni x$ be a compact neighborhood of the point $x$ in the space $A$ and $V\ni x$ be a compact neighborhood of the point $x$ in the space $B$. Then $U\cap V$ is a neighborhood of the point $x$ in the space $A\cap B$. Since the space $X$ is Hausdorff, both sets $U$ and $V$ are closed in the space $X$. Hence $U\cap V$ is a closed subset of the space $X$ too. Then $U\cap V$ is a closed subsets of a Hausdorff compact space $U$ too, so a space $U\cap V$ is compact.
• Could you explain why are $U$ and $V$ closed in $X$? $U$ is a compact neighborhood of the locally compact Hausdorff subspace $A$, so it is closed in $A$. But does that mean it is closed in $X$? – eatfood Mar 27 at 4:22
• @eatfood Every compact subset of a Hausdorff space $X$ is closed in $X$, see, for instance, Engelking’s “General topology”, Theorems 3.1.8 and 3.1.6. – Alex Ravsky Apr 2 at 19:54