4
$\begingroup$

Fact

The intersection of a closed set $A$ and a compact set $B$ is closed.

Background

I wanted to prove that the intersection of closed and compact sets is compact. There might be other ways, but I am not interested in them for now and am looking to prove the above statement.

Relevant Facts

I don't have much idea what to use. $A^c$ is open. Every cover $\cup_{i\ge1}U_i$ of $B$ contains a subcover $\cup_{i=1}^kU_i$. I might want to prove $A\cap B^c$ is open. Limit points may be another points to consider but I don't get any idea in that direction.

I also don't have that $B$ is closed otherwise it would be trivial.

Some Clarification

I am aware that compact sets are not closed in arbitrary topological spaces. The reason why I was motivated to believe tha intersection of closed and compact set is compact is true is: See Matt E.'s Answer Here.

So if $K$ is compact in an arbitrary topological space (which is just to say that it is a compact topological space when given its induced topology) and $L$ is closed then $K \cap L$ is a closed subset of $K$ in its induced topology

Is it true if $A$ and $B$ are not whole spaces?

I would appreciate any hints.

$\endgroup$
  • 2
    $\begingroup$ Are there supposed to be subsets of $\mathbb{R}$, or of an arbitrary topological space? $\endgroup$ – Zev Chonoles May 29 '13 at 22:58
  • $\begingroup$ If you aren't assuming enough to prove that compact sets are closed, then the "fact" can fail. Maybe $A$ is the whole space, and $B$ is a compact set that is not closed. Are you working in a Hausdorff space? $\endgroup$ – Trevor Wilson May 29 '13 at 23:01
  • $\begingroup$ Also, it seems you are in a context where "countably compact" and "compact" are equivalent. Is this correct? $\endgroup$ – Trevor Wilson May 29 '13 at 23:03
  • $\begingroup$ @MinimusHeximus How does that relate to the problem? We have one compact set and one closed set. $\endgroup$ – Trevor Wilson May 29 '13 at 23:17
  • 1
    $\begingroup$ @user45099 You have misunderstood Matt E's post. If $K$ is any subset whatsoever and $L$ is closed, then $K \cap L$ is closed as a subset of $K$. $\endgroup$ – Zhen Lin May 30 '13 at 21:03
6
$\begingroup$

It is not always true. The following example witnesses it.

Example 1: Let $X= \mathbb R$ with finite complement topology. Note that $X$ is $T_1$, not $T_2$. It is not difficult to prove that $X$ is compact. Now let $A=X$; $B=\{0,1,2,\dots,n,\dots\}$. Note that $A$ of course is closed and $B$ is compact, however it is not closed. Therefore $A \cap B=B$ is not closed.

However it is true if $A$ and $B$ belong to a Hausdorff space.

Theorem 2: Every compact subspace of a Hausdorff space $X$ is a closed subspace of $X$.

Proof: Let $A$ be a compact subspace of $X$. For every $x\in X\setminus A$ there exists an open set $V\subset X$ such that $x \in V$ and $A \cap V=\emptyset$, so that $X \setminus A$ is an open subset of $X$.

So in a Hausdorff space, $B$ is always closed as it is compact, and hence $A \cap B$ is closed, as the intersection of two closed set is closed.

$\endgroup$
  • $\begingroup$ Your examples is convincing. $\endgroup$ – user45099 May 30 '13 at 5:36
4
$\begingroup$

I believe this constitutes a minimal counterexample to the modified question. Note that it is not a very "nice" space—it is neither $T_0$ nor preregular—but as requested, $A$ and $B$ are proper subsets of $X$:

Let $X = \{ 1, 2, 3 \}$.

Let $\tau = \{\varnothing, \{1\}, X \}$ be the topology for $X$.

Let $A = \{ 2, 3 \}$ and let $B = \{2\}$.

Then $A$ is closed because its complement is open, $B$ is compact because every finite set is compact, but their intersection, $\{2\}$, is not closed.

$\endgroup$
0
$\begingroup$

That the intersection of a closed set with a compact set is compact is not always true. However, if you further require that the compact set is closed, then its intersection with a closed set is compact.

First, note that a closed subset $A$ of a compact set $B$ is compact: let $U_i$, $i \in I$, be an open cover of $A$; as $A$ is closed, $B \setminus A$ is open and so $U_i \cup (B \setminus A)$ is an open cover of $B$. From the compactness of $B$ we deduce that there exists a finite subcover of open sets $U_j$ for $A$, which is therefore compact.

Now, let $U$ be a compact closed set and $V$ be a closed set of a topological space $X$. As both $U$ and $V$ are closed, the intersection $U \cap V$ is closed; as $U$ is compact and $U \cap V \subseteq U $, by the previous remark, it follows that $ U \cap V$ is compact.

Note that for the claim to be true it is not required that $X$ is Hausdorff, but merely that the compact set $U$ is also closed. Indeed, as in a Hausdorff space every compact set is closed, the claim that the intersection of a closed with a compact is compact is true as a corollary of the more general case of which I have here given a proof.

$\endgroup$

Your Answer

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