September 2001 #5

Problem Statement: Use the open cover definition of compact to show that E=\{0\}\cup\{1,1/2,1/3,1/4,\dots\} is a compact subset of \mathbb{R}.

Proof: Let S be an open cover of E, S=\bigcup_{j\in\mathcal{J}}S_j for some indexing set \mathcal{J}. If S is a finite cover then we’re done, so suppose that S is not a finite open cover of E.

Since 0\in E there is at least one S_k\in S such that 0\in S_k. Since \{1/n\} converges to 0 there is an N\in \mathbb{N} such that 1/n\in S_k for every n>N. Furthermore, since S is an open cover of E there is at least one S_j\in S such that 1\in S_j. Pick one of these and call it S_1. We may do this for 1/2, 1/3,\dots,1/N and find S_2,S_3,\dots,S_N.

Let T=S_k\cup S_1\cup S_2\cup\dots\cup S_N which is contained in S. We claim that T is a finite open cover of E. First note that each of the S_n are open and so their union is open. Let x=\dfrac{1}{n}\in E. If n>N then x\in S_k, otherwise, x\in S_n. Either way, x\in T. Thus, T is a finite open cover of E.

Since such a T can be constructed from any open cover S of E it follows that every open cover of E in \mathbb{R} has a finite subcover in \mathbb{R}, thus, E is compact in \mathbb{R}.

\Box

Reflection: The most important part of this proof rests in the fact that 0 is the limit of the sequence \{\dfrac{1}{n}\}. This allows us to cover infinitely many of the points in E with a single open set S_k and leave only finitely many left to cover. After that, the rest is a snap 🙂

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This entry was posted in Analysis, Compact, Sequence and tagged , , . Bookmark the permalink.

4 Responses to September 2001 #5

  1. j2kun says:

    Ooooooooh snap!

  2. Sir Wadesworth says:

    Wow, someone knows how to spend their sunday nights!

  3. DR says:

    We’re the ones reading math blogs on Sunday, Wadesworth!

  4. Michael says:

    Squeaky clean.

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