I need to use both the sequential definition of compactness and the Heine-Boral theorem (i.e. provide to separate proofs) to prove the following statement

the union of finitely many compact subsets of $\mathbb{R}^n$ must be compact.

More Info:

  • Definition of compactness A subset $S$ of a normed vector space $V$ is compact if and only if every sequence in $S$ has a subsequence which converges to some vector in $S$
  • Heine-Borel Theorem: The compact subsets of $(\mathbb{R}^n,\Vert \Vert)$ are the closed and bounded subsets of $\mathbb{R}^n$.


Using Definition: Let $\{x_n : n=1,2,3,4...,i \}$ be sets which satisfy the Heine-Borel Theorem (they are bounded and closed). Let $X=\cap^i_{n=0}x_n$. Let $a$ be a limit point of $X$. Then there must be exist a sequence $\{b_i, i \in \mathbb{N}\}$ in which $X$ converges to $a$. We need to show that $a \in X$. Suppose $a \not \in X$. Let $ \epsilon = min(\Vert a-x_n \Vert: n=1...i) $. Since $\lim_{n \rightarrow \infty} b_i=a$ there exists $N$ such $n>N$ implies $\Vert a-b_i\Vert < \epsilon/2$. Since $\Vert x_n-a\Vert \geq \epsilon$ for all n=1to i we have a contradiction. Then $a$ must belong to $X$. Hence $X$ is closed. (not sure how to extend this to compactness).

Using Heine-Borel Theorem:Very confused not sure what to do.


Is it true that the finite union of closed sets is closed in $\mathbb{R}^n$? If we have finitely many bounded sets, can we find a bound for their union - that is; can we find a disk centered at the origin which contains all of these sets?

The answer is yes to both: by definition the union of closed sets is closed, so that criterion is met. Also, the finite union of bounded sets is bounded, because each bounded set is contained in a disk of finite radius centered at the origin. We can take the maximum of the radii called $R_{\text{max}}$ of these disks to see that the union of these sets is contained in $B(0,R_{\text{max}})$.

So, such a finite union is closed and bounded, and by the Heine-Borel Theorem is compact.

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