6

I've been stuck on the following problem for several days: Let $(M,d)$ be an arbitrary metric space and $S, T$ be subsets of $M$.

If $S$ is closed and $T$ is compact, then $S \cap T$ is compact.

I know that if $T$ is compact, $T$ is closed and bounded. That would imply that $S \cap T$ is also closed and bounded since $(S \cap T) \subseteq T$. Also since $S$ is closed, $S$ contains all its accumulation points.

Other than writing down the definitions, I really don't know how to proceed. Could someone give me a hint?

Davide Giraudo
  • 172,925
Student
  • 3,353
  • What definition of compactness do you use? – Stefan Hamcke Aug 15 '13 at 21:27
  • 1
    @StefanH.: My book states that a subset $S$ of a metric space $M$ is called compact if every open covering of $S$ contains a finite subcover. – Student Aug 15 '13 at 21:28
  • 6
    Work directly with the definition of compactness. "Closed and bounded" fails in general metric spaces. – André Nicolas Aug 15 '13 at 21:32
  • 2
    @AndréNicolas: I think another theorem in my book helps me solve the problem. It says that if $X$ is a closed subset of a compact metric space $M$, then $X$ is compact. Since $T$ is compact, $S\cap T$ is closed and $S\cap T \subseteq T$, $S \cap T$ is compact. – Student Aug 15 '13 at 21:36
  • @CameronBuie: Thanks! I'm still trying to solve the problem another way with your hint though. – Student Aug 15 '13 at 21:42

3 Answers3

3

Hint: Let $\mathcal{C}$ be an open cover of $S\cap T.$ Note that $\mathcal{C}\cup\{M\setminus S\}$ (where $M\setminus S$ denotes the complement of $S$ in $M$) is an open cover of $T$. (Why?) Can you take it from there?

Alternately, as you (astutely) observed, in the comments, you can simply use the Theorem that says a closed subset of a compact metric space is compact, noting that $T$ is a compact metric space and that $S\cap T$ is a closed subset of $T$.

Cameron Buie
  • 102,994
  • If $x \in T$, then certainly $x \in \mathcal{C} \cup {M \setminus S }$. And since the union of two open sets is open, $\mathcal{C} \cup {M \setminus S }$ is an open cover for $T$. Since $T$ is compact, there exists a finite subcover of $\mathcal{C} \cup {M \setminus S }$ that covers $T$. Since $(S \cap T) \subseteq T$, $\mathcal{C} \cup {M \setminus S }$ must also have a finite subcover for $S \cap T$. – Student Aug 15 '13 at 21:49
  • I think I can make the claim that since ${ M \setminus S }$ and $S \cap T$ are disjoint, that there exists a finite subcover of $\mathcal{C}$ that covers $S \cap T$. So we would be done – Student Aug 15 '13 at 21:56
  • You are right on target, except for one thing: $\mathcal{C}$ and ${M\setminus S}$ are sets of open sets ($M\setminus S$ is open because $S$ is closed, of course), not open sets themselves. What you need to do is show that $$T\subseteq(M\setminus S)\cup\left(\bigcup_{U\in\mathcal{C}}U\right),$$ whence you will have shown that $\mathcal{C}\cup{M\setminus S}$ is an open cover of $T.$ – Cameron Buie Aug 15 '13 at 22:11
1

To use your definition of compactness, consider an open covering of $S$. Augment it with the complement of $S$, which is open. This gives you an open covering of $T$. Fill the details and finish the proof.

lhf
  • 216,483
0

$S\cap T$ is a closed subset of $T$ (relative to $T$, closedness of $T$ is actually not needed). It is a general fact in topology that a closed subset of a compact space is compact. To show that, let $X$ be a compact topological space (or a metric space), $A$ a closed subset of $X$, and $\mathcal U=\{U_i\mid i\in I\}$ an open cover of $A$. Now consider the open cover $\mathcal U \cup\{X-A\}$ of $X$.

Stefan Hamcke
  • 27,733