Does anyone have a proof that the intersection and union of two compact sets is compact.

Question

I have my take on it. It is quite informal and don;t know where it would be evaluated correctly on an exam. Since the sets are compact that means for every open cover there is a finite cover.

When the intersection is in question, taking any two covers of the sets, then there respective finite covers and taking that intersection of two finite covers gives us a finite cover of an intersection.

For union just taking the finite covers and uniting them also gives us a finite cover of a union. Would this be sufficient in your opinions?

Answer 1

For union you are right.

(For Hausdorff spaces only) For intersection you can think in this way too take any open cover of $A\cap B$ then extend it to an open cover of $A$ or $B$, say $A$ [by taking $x\in A-B$ and choose a fixed $z\in A\cap B$ then $\forall x\in A-B$ take open sets $U_x$ now take their union, then $A$ has finite subcover and that corresponding subcover will also be the subcover of $A\cap B$.

Answer 2

Since you used calculus as a tag, I will assume for this question that you are using the Euclidean Space, that is, any subet of $\mathbb R^n$.

I will provide you with a proof for the union.

Let $A$ and $B$ be any compact subsets of $\mathbb R^n$.

Since $A$ and $B$ are compact, we know from Heine-Borel Theorem that they are closed and bounded, that is $\exists M_1, M_2 \in \mathbb R^n$ such that $a \le M_1$ for all $a \in A$ and $b \le M_2$ for all $b \in B$. Now, choose $M^*= \max\{M_1, M_2\}$ then we clearly have that $x \le M^*$ for every $x \in A \cup B$. Thus $A\cup B$ is bounded.

We must now show that $A \cup B$ is closed and then we are done. Notice, however, that $A \cup B$ is the union of a finite number of closed sets and must thus be closed.

We thus have that $A \cup B$ is compact in $\mathbb R^n$, since it is both closed and bounded in $\mathbb R^n$.