Show that the union of finitely many compact sets is compact

Question

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.

Answer 1

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.

Answer 2

Assume that $|x|\leq M_{i}$ for all $x\in A_{i}$, $i=1,...,n$, then $|x|\leq\max_{1\leq i\leq n}M_{i}$ for all $x$ in the union.

Assume that $x_{n}$ is in the union with such that $x_{n}\rightarrow x$, then at least one $A_{i}$ must contain infinitely many $x_{n}$, say a subsequence $x_{n_{k}}$. As $A_{i}$ is closed, then $x_{n_{k}}\rightarrow x\in A_{i}$.

Answer 3

Let $C_1,...,C_n$ be compact, there exists $R_i$ such that $C_i\subset B(),R_i)$, let $R=Sup(R_i), C_1\cup...\cup C_n\subset B(0,R)$

Let $x$ be a limit point of $C=C_i\cup...C_n$, there exists $(x_i)$ such that $x_i\in C$ and for every $e>0$, there exists $N$ such that $i>N$ implies $d(x,x_i)<e$. There exists $i_0$ and a subsequence $x_{i_p}\in C_{i_0}$ of $(x_i)$, $x=limx_{i_p}$ since $C_{i_0}$ is closed, $x\in C_{i_0}\subset C$.

Answer 4

Assume $K_j, j=1\cdots n$ are compact sets and

Note that$$\overline{A\cup B}=\overline{A}\cup\overline{ B} $$ and hence, for finite union and since each $K_j$ is closed we have, $$\overline{\bigcup_{j=1}^{n}K_j}=\bigcup_{j=1}^{n}\overline{K_j } =\bigcup_{j=1}^{n} K_j $$ On the other hand, all each $K_j$ is bounded hence there is $R_j>0$ such that

$$K_j\subset B(0, R_j)\subset B(0, \max(R_j)) $$ thus $$\bigcup_{j=1}^{n}K_j\subset B(0, R) ~~~~~~R=\max(R_j)$$

this prove that $\bigcup_{j=1}^{n}K_j$ is bounded and closed hence compact.

Alternatively, in topological way you have,

Let $(O_i)_{i\in I}$ be any open covering of $\bigcup_{j=1}^{n}K_j$ then it is also a covering of the compacts $K_j$ Thus by compactness of $K_j$ there is $i_1\cdots i_j\in I$ such that $$K_j\subset \bigcup_{k=1}^{j}O_{i_k}$$

this implies $$\bigcup_{j=1}^{n}K_j\subset \bigcup_{j=1}^{n}\bigcup_{k=1}^{j}O_{i_k}$$

Which prove the compactness of $\bigcup_{j=1}^{n}K_j$ since $(O_{i_k})_{1≤k≤j\\1≤j≤n}$ is thereof a finite covering.