Let (X, d) be a metric space and A, B ⊂ X be two compact subsets. Show that A ∩ B is also compact

Question

Question seems fine i just have a few doubts.

Is it possible to just use the Heine Borel theorem? as both A and B are compact it implies they are both closed, so therefore their intersection is closed.

I'm just not too sure how to go about showing boundedness. Can you say A ∩ B ⊂ A and A ∩ B ⊂ B where A is bounded above and below by a and b resp. and B is bounded above and below by c and d resp. which implies b is a lower bound and c is an upper bound for A ∩ B ? Therefore it is bounded and compactness follows.

Answer 1

You could not use boundedness to prove compactness in general metric spaces. Take any infinite set $M$ with the metric $d(x,y)=1$ if $x\not=y$. This makes $M$ a metric and a topological space with the discrete topology, every point is an open set (since $B(x,1/2)=\{x\}$ for every $x\in M$). Then the open cover $\{\{x\}: x\in M\}$ has no finite subcover, hence $M$ is not compact. Nevertheless $M\subseteq B(x,2)$ (any $x$), thus $M$ is bounded.

To prove that $A\cap B$ is compact in general (if both are compact subsets of a metric space $X$), take any finite open cover of $A\cap B$ and add to it the set $X\setminus(A\cap B)$ (which is open) so you get a finite cover of $A$ (and also of $B$) so now take a finite subcover of this, using that $A$ is compact. Then throw out the set $X\setminus(A\cap B)$ (if it is in your finite subcover of $A$), what remains is a finite subcover of $A\cap B$.

You didn't ask about it, but you might try to use this idea to prove that $A\cup B$ is also compact.

Answer 2

Hint: in any metric space, any closed subset of a compact set is compact.

Answer 3

Since $A$, $B$ are compact sets, we deduce $A$, $B$ are the close sets. So, $A\cap B$ is also close set. Let the sequence $(x_n)\subset A\cap B$. This implies $(x_n)\subset A$. But $A$ is compact, we infer that there exists a subsequence $(x_{n_k})$ of $(x_n)$ such that $x_{n_k}\to a\in A$ as $k\to\infty$. Noting that $(x_{n_k})\subset A\cap B$ and $A\cap B$ is close, we get $a \in A\cap B$. That means that $A\cap B$ is compact.