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

  • Thread starter ArcanaNoir
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  • #1
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Homework Statement



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].


Homework 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.

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]
 
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Answers and Replies

  • #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)))?
 
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  • #3
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int(A) is an open set contained in cl(int(A)). What does this tell you about it's relation to int(cl(int(A)))?
[itex] \mathring{A}\subseteq \mathring{\bar{\mathring{A}}} [/itex]
 
  • #4
Dick
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[itex] \mathring{A}\subseteq \mathring{\bar{\mathring{A}}} [/itex]
Ok, so it's pretty easy to finish from there, right?
 
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  • #5
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oh! Thank you :)
 

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