Show that the union of finitely many compact sets is compact.
Note: I do not have the topological definition of finite subcovers at my disposal. At least it wasn't mentioned. All I have with regards to sets being compact is that they are closed and bounded by the following definitions:
Defn: A set is closed if it contains all of its limit points
Defn: A set is bounded if $\exists$ R such that the set $A$ is contained in the $B_{R}(0)$
Attempt:
Suppose $\bigcup_{i = 1}^nA_{i}$ is not compact. $$\Rightarrow \exists\ A_{i} \ such\ that \ A_{i}\ is\ not\ compact. $$
But we assumed each of the individual $A_{i}$ were compact. Therefore a contradiction.
Best Answer
Hints:
(1) Can you show that a finite union of closed sets is closed?
(2) Can you show that a finite union of bounded sets is bounded?
(3) Do you see that (1) and (2) mean that a finite union of closed and bounded sets is closed and bounded?
Additional Hints:
(1) is equivalent to showing that a finite intersection of open sets is open
For (2), consider the fact that $\max\{B_1,B_2,\ldots,B_N\} \geq B_k$ for each $k$, so the largest bound is a common bound for all of the sets.