8
$\begingroup$

Let X be a locally compact topological space. I need to prove that if $M\subset X$ is a locally compact subspace of X then there exist $U,F\subset X$ such that U is open and F is closed, and $M=U\cap F$. It can be assumed that for every open $x\in V$ there exist an open subset U such that $x \in U \subset \overline U \subset V$ and $\overline U$ is compact.

My thoughts about the problem are that for every $x\in M$ and for every neighborhood $x\in V_x$ there exist an open neighborhood $U_x$ in M such that $x\in U_x\subset \overline U_x \subset V_x$ and $\overline U_x$ is compact and closed in M. So there exist an open subset $Z_x\subset X$ such that $U_x = Z_x\cap M$. I can take $U = \bigcup _{x\in M}Z_x $, then U is open in X, but i was having trouble finding the closed set F. I thought to take $F = \overline {\bigcup _{x\in M}U_x}$, that is closure in X. Now $M\subset U\cap F$, but I haven't succeeded in showing that $M\supset U\cap F$, maybe that's not even true.

Thanks!

$\endgroup$
7
$\begingroup$

Without assuming that $X$ is Hausdorff (and using the definition of local compactness given) the result is not true.

Let $X$ be any set with at least two elements, and consider the trivial (anti-discrete) topology on $X$. Clearly this space has the property that the nonempty open sets with compact closures form a base (indeed, there is only one nonempty open set, and it's closure is clearly compact), and so this space is locally compact. Note, too, that any nonempty proper $M \subseteq X$ is also locally compact (another anti-discrete space), but will not be the intersection of an open and a closed subset of $X$.


However, the result does follow if we assume that $X$ is Hausdorff. (Then local compactness is equivalent to every point having an open neighbourhood with compact closure.)

To wit: given $x \in M$, let $V_x$ be a neighbourhood of $x$ in $M$ such that $\mathrm{cl}_M ( V_x ) = \overline{V_x} \cap M$ is compact. Now $\overline{V_x} \cap M$ is also a compact subset of $X$, so by Hausdorffness $\overline{V_x} \cap M$ is closed (in $X$). Fix an open $W_x \subseteq X$ such that $W_x \cap M = V_x$. Clearly we have that $$\overline{ W_x \cap M } \cap M = \mathrm{cl}_M (V_x).$$ Note, now, that as $W_x \cap M \subseteq \overline{ W_x \cap M } \cap M$, we have that $$ W_x \cap \overline{M} \subseteq \overline{ W_x \cap \overline{M} } = \overline{ W_x \cap M } \subseteq \overline{ W_x \cap M } \cap M \subseteq M, $$ (and clearly $x \in W_x \cap \overline{M}$).

Setting $U = \bigcup_{x \in M} W_x$, it follows that $U$ is open in $X$, and $M = U \cap \overline{M}$, as desired.

$\endgroup$
  • $\begingroup$ I don't see how you got that $\overline{W_x\cap M}\subset\overline{W_x\cap M}\cap M$. How did you get that? $\endgroup$ – Anonymous Jul 25 '18 at 17:46
2
$\begingroup$

Let us assume that $X$ is a KC space, that is a space where all compact sets are closed (In particular, each Hausdorff space is KC). Then $M$, as a subspace, is also KC. Now it suffices if every point in $M$ has a compact neighborhood, as then we can assume that this compact neighborhood is the closure of some open neighborhood. With these hypotheses, let's show that $M$ is the intersection of an open and a closed subset of $X$:
Each $x\in M$ has an open neighborhood $N$ in $M$, whose closure relative to $M$ is compact and a subset of $M$. But that implies, as $X$ is KC, that this relative closure is closed in $X$, hence $$\overline N^M=\overline N\subseteq M$$ Since $N$ was open in $M$, there is an open $N'$ such that $N=M\cap N'$. Now $$\overline M∩N'=\overline{M∩N'}^{N'}=\overline N^{N'}=\overline N∩N'\subseteq M∩N'=N$$ This means that $N=\overline M∩N'$ is also a neighborhood of $x$ in $\overline M$. Since $x\in M$ was arbitrary, and $N\subseteq M$, this shows that $M$ is open in $\overline M$. In particular, $M$ is the intersection of some open set with $\overline M$.


Edit (06/12/2015): The proof becomes shorter using the notion of local closedness and assuming that you have proven that a locally closed subset is open in its own closure. Here a set $Y$ of a space is locally closed if each $y\in Y$ has a neighborhood $U$ such that $Y\cap U$ is closed in $U$.
Now if $Y$ is a locally compact subspace of the KC space $X$, then each $y\in Y$ has a neighborhood $K'$ in $X$ such that the relative neighborhood $K=Y\cap K'$ is compact. But that means that $Y\cap K'$ is closed, so $Y$ is locally closed.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.