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Thread: Proving something is compact

  1. #1
    Junior Member
    Mar 2009
    Madison, WI

    Proving something is compact

    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}.
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  2. #2
    Member HappyJoe's Avatar
    Sep 2010
    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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