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Thread: Proving something is compact

  1. #1
    Junior Member
    Mar 2009
    Madison, WI

    Proving something is compact

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

    Definition from the book:
    A subset $\displaystyle K$ of a metric space $\displaystyle X$ is said to be compact if every open cover of $\displaystyle K$ contains a finite subcover. More explicitly, the requirement is that if $\displaystyle \{G_{\alpha}\}$ is an open cover of $\displaystyle K$, then there are finitely many indices $\displaystyle \alpha_1, ... \alpha_n$ such that $\displaystyle K \subset G_{\alpha_1} \bigcup ... \bigcup G_{\alpha_n}.$
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  2. #2
    Member HappyJoe's Avatar
    Sep 2010
    I cannot resist first telling you the easy method:

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

    Maybe you already thought of this.

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

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

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

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

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