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Prove: If p has a neighborhood contained in A, then p is in the interior of A.

  1. Sep 3, 2012 #1
    1. The problem statement, all variables and given/known data
    If p has a neighborhood contained in A, then p is in the interior of A.


    2. Relevant equations
    Int A = [itex]\bigcup[/itex]{C[itex]\subseteq[/itex]X:C[itex]\subseteq[/itex]A and C is open in X}
    By the books definition, a neighborhood is open.

    3. The attempt at a solution
    Let C'[itex]\subseteq[/itex]A be a neighborhood of p contained in 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}
    p[itex]\in[/itex]Int A

    Seems too simple, but I'm going through the book's definition of "neighborhood of p"--and open subset of X.
     
    Last edited: Sep 3, 2012
  2. jcsd
  3. Sep 3, 2012 #2
    You are thinking along the right lines. But C' as you chose it leads you nowhere. You have not proved that p is in any of C.
     
  4. Sep 3, 2012 #3
    Should it state something like:
    Let p[itex]\in[/itex]A,
    There exists an open set C'[itex]\subseteq[/itex]A s.t. p[itex]\in[/itex]C'
    Thus it's in the union of all open sets.

    Would that fix the issue? Thanks!
     
  5. Sep 3, 2012 #4
    That works.
     
  6. Sep 3, 2012 #5
    Excellent! Thank you for the help!
     
  7. Sep 3, 2012 #6
    You just need to put things in a logical sequence without any gaps. First there is a neighborhood V. It contains some open set C'. Which must necessarily be one of all those open sets. So p is in the union. Q. e. d. If you are not explicit about every small detail, your professor might become unhappy :)
     
  8. Sep 3, 2012 #7
    Lol, my professor is a stickler--but she is an excellent teacher! So she's only a stickler because she has covered the material. So hopefully I catch these things before a test---though we have a quiz coming up!

    If you have time, do you mind helping me with my other post? I'm not quite sure I can stop where I did--though it makes sense to me!
     
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