Proving something is compact

**Zennie** · September 29th 2010, 07:00 PM

Let $K \subset R^1$ consist of $0$ and the numbers $1/n$ for any $n = 1, 2, 3 ...$ Prove that $K$ is compact directly from the definition (w/o using the Heine-Borel Theorem).

Definition from the book:
A subset $K$ of a metric space $X$ is said to be compact if every open cover of $K$ contains a finite subcover. More explicitly, the requirement is that if $\{G_{\alpha}\}$ is an open cover of $K$ , then there are finitely many indices $\alpha_1, ... \alpha_n$ such that $K \subset G_{\alpha_1} \bigcup ... \bigcup G_{\alpha_n}.$

**HappyJoe** · September 29th 2010, 11:14 PM

I cannot resist first telling you the easy method:

Let $A = \{\frac{1}{n}|n\in\mathbb{N}\}\subset\mathbb{R}$ . Then the closure of A is equal to your set $K$ , whereas $K$ is closed. Clearly, $K$ is also bounded, which is why $K$ is compact.

Maybe you already thought of this.

The other (and in this case legal) way: Let $\{G_\alpha\}$ be an open cover of $K$ . In particular, $0\in K$ is contained in some $G_\alpha$ , say $0\in G_{\alpha_0}$ .

An open subset of $\mathbb{R}$ is a disjoint union of open intervals. In particular, the point 0 is contained in an open interval inside of $G_{\alpha_0}$ . Suppose this open interval has $\delta$ as its right end point. Then all points of $K$ with $\frac{1}{n}<\delta$ are also contained in this interval, i.e. all such points are contained in $G_{\alpha_0}$ . But all but finitely many points $1,\frac{1}{2},\frac{1}{3},\ldots,\frac{1}{k}$ of $K$ satisfy $\frac{1}{n} < \delta$ , and so all but finitely many points of $K$ are contained in $G_{\alpha_0}$ .

Since we have a cover of the set $K$ , the point $\frac{1}{j}$ is contained in some set $G_{\alpha_j}$ , for each $j=1,2,\ldots,k$ . But then the cover

$\{G_{\alpha_0}, G_{\alpha_1},G_{\alpha_2},\ldots,G_{\alpha_k}\}$

is a finite subcover of $\{G_\alpha\}$ .

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