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Basic topology: closure and union

  1. Mar 27, 2012 #1
    1. The problem statement, all variables and given/known data
    [itex]Cl(S \cup T)= Cl(S) \cup Cl(T)[/itex]


    2. Relevant equations
    I'm using the fact that the closure of a set is equal to itself union its limit points.


    3. The attempt at a solution
    I am just having trouble with showing [itex]Cl(S \cup T) \subset Cl(S) \cup Cl(T)[/itex]. I can prove this one way, but I figured out another way that I prefer and want to know if there is any flaw in it:

    Basically it boils down to showing the limit points of [itex]S \cup T[/itex] is a subset of the limit points of S union the limit points of T. So, if x is a limit point of [itex]S \cup T[/itex] then every open set U containing x intersects [itex]S \cup T[/itex] in a point other than x. Hence, x is a limit point of S or T.

    I keep going back and forth. Sometimes I feel like this is fine and other times I feel like I'm making an error because if x is a limit point of S this means that every neighborhood of x intersects x in a point other than x. But in the proof I just gave, we know every neighborhood of x intersects [itex]S \cup T[/itex] in a point other than x, so we don't know if it necessarily always in S or T. I don't know why I'm having so much trouble with this! I think I am just over thinking it or something. So is this proof fine then?
     
    Last edited: Mar 27, 2012
  2. jcsd
  3. Mar 27, 2012 #2

    Dick

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    Well, I think you've figured out exactly what the problem is. And it is a problem. It's good to have a feeling for when a proof is incomplete. You haven't proved it this way.
     
  4. Mar 28, 2012 #3
    I'm not certain this can be proved directly like this, but the contrapositive is pretty easy to show. If x is not in Cl(S)∪Cl(T), then show it can't be in Cl(S∪T).
     
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