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Show If a point A is in the interior, then it has a neighborhood contained in A.

  1. Sep 2, 2012 #1
    1. The problem statement, all variables and given/known data
    Let A be a topological space and let A[itex]\subseteq[/itex]X be any subset.
    Show: If a point A is in the interior, then it has a neighborhood contained in A.


    2. Relevant equations
    Neighborhoods are defined to be open in my book.
    Int(A) = [itex]\bigcup[/itex]{C[itex]\subseteq[/itex]A and C is open in X}


    3. The attempt at a solution
    Let p[itex]\in[/itex]Int(A).
    Then p[itex]\in[/itex][itex]\bigcup[/itex]{C[itex]\subseteq[/itex]X:C[itex]\subseteq[/itex]A and C is open in X}
    So, [itex]\exists[/itex] an open set C' s.t. p[itex]\in[/itex]C' and C[itex]\subseteq[/itex]A
    Q.E.D.
     
  2. jcsd
  3. Sep 2, 2012 #2

    micromass

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    Sounds good!
     
  4. Sep 2, 2012 #3
    Thank you! Could you check my work backwards now?
    I can say:

    Let C'[itex]\subseteq[/itex]A be open and non-empty
    Let p[itex]\in[/itex]C
    Thus p[itex]\in[/itex][itex]\bigcup[/itex]{C[itex]\subseteq[/itex]X:X[itex]\subseteq[/itex]A and C is open in X}
    Thus p[itex]\in[/itex]Int A
     
  5. Sep 2, 2012 #4

    micromass

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    That's correct. But you got to clean up your presentation. What's C' for example??

     
  6. Sep 2, 2012 #5
    C' was meant to be one set that has that element and
    I meant to type let p[itex]\in[/itex]C'
    and I meant to write
    p∈⋃{C⊆X:C⊆A and C is open in X}

    Do I still need to clean up?
     
  7. Sep 2, 2012 #6

    micromass

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    Ah ok. I guess it's ok now!
     
  8. Sep 2, 2012 #7
    Thank you very much! If you have time, would you mind helping me with my other question? It's about the equivalence of a bounded set with a closed ball. I just want to check my proof one direction. Thanks again!
     
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