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Prove cl(int(cl(int(A))))=cl(int(A))

  1. Oct 19, 2013 #1
    1. The problem statement, all variables and given/known data

    I am working on the proof that taking closure and interior of a set in a metric space can produce at most 7 sets. The piece I need is that [itex] \bar{\mathring{A}} = \bar{\mathring{\bar{\mathring{A}}}} [/itex].


    2. Relevant equations

    Interior of A is the union of all open sets contained in A, aka the largest open set contained in A.
    Closure of A is the intersection of all closed sets containing A, aka the smallest closed set containing A.

    3. The attempt at a solution

    [itex] \bar{\mathring{A}}[/itex] is a closed set. [itex]\mathring{\bar{\mathring{A}}}\subseteq \bar{\mathring{A}}[/itex]. Since [itex]\bar{\mathring{\bar{\mathring{A}}}} [/itex] is the smallest closed set containing [itex] \mathring{\bar{\mathring{A}}} [/itex] we have that [itex]\bar{\mathring{\bar{\mathring{A}}}}\subseteq \bar{\mathring{A}}[/itex].

    I'm not sure how to get the inclusion [itex] \bar{\mathring{A}}\subseteq \bar{\mathring{\bar{\mathring{A}}}} [/itex]
     
    Last edited: Oct 19, 2013
  2. jcsd
  3. Oct 19, 2013 #2

    Dick

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    int(A) is an open set contained in cl(int(A)). What does this tell you about its relation to int(cl(int(A)))?
     
    Last edited: Oct 19, 2013
  4. Oct 19, 2013 #3
    [itex] \mathring{A}\subseteq \mathring{\bar{\mathring{A}}} [/itex]
     
  5. Oct 19, 2013 #4

    Dick

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    Ok, so it's pretty easy to finish from there, right?
     
  6. Oct 19, 2013 #5
    oh! Thank you :)
     
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