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Close of a Singleton in R1

  1. Mar 10, 2013 #1
    Let me add some clarification here, since I realized that my notation might be confusing. By [itex]\left\{x\right\}'[/itex] below, I mean the set of all limit points of [itex]\left\{x\right\}[/itex]. Hope this helps anyone trying to figure out what I mean, thanks!

    The title says R1, but it's actually just a metric space [itex]X[/itex], my apologies.

    1. The problem statement, all variables and given/known data

    Let [itex]X[/itex] be a metric space and let [itex]x\in{X}[/itex] be any point. Prove that the set [itex]\left\{x\right\}[/itex] is closed in [itex]X[/itex]


    2. Relevant equations

    NA

    3. The attempt at a solution

    Ok, so I realize I can prove this by showing that the complement is open (simple). However, I have a question about the validity of an alternative approach:

    It is clear that [itex]\left\{x\right\}[/itex] has no limit points, thus [itex]\left\{x\right\}' = [/itex]empty set. Now, the empty set is a subset of every set, thus [itex]\left\{x\right\}'\subset{\left\{x\right\}}[/itex] and so [itex]\left\{x\right\}[/itex] contains all of its limit points. Then [itex]\left\{x\right\}[/itex] is closed.

    Thoughts? Does this work? If not, what am I missing? Thanks!
     
    Last edited: Mar 10, 2013
  2. jcsd
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