Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Set closure and interior points

  1. Mar 3, 2005 #1
    If [tex]A \subset X[/tex] where [tex]X[/tex] has a topology, is it generally true that the interior of [tex]A[/tex] is equal to the interior of the closure of [tex]A[/tex]? This seems very reasonable to me, but probably only because I'm visualizing [tex]A[/tex] as a disc in the real plane. If it isn't true, what would be a counterexample?

    thanks
     
  2. jcsd
  3. Mar 3, 2005 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Take X to be the set of real numbers with the usual topology.

    Let A be the set of all rational numbers between 0 and 1 (inclusive).

    What is the interior of A? What is the closure of A?

    What is the interior of the closure of A?
     
  4. Mar 3, 2005 #3
    Okay this is a good example! Let me see if I have it straight. The interior of A is the largest open set contained in A, which would be the empty set in this case. I say this because the set has no interior points, or points for which a neighborhood can be found containing only points in A. The closure of A is the smallest closed set containing A, which is the interval [0, 1]. I say this because for every point in [0, 1], every neighborhood of that point contains points in A. But the interior of [0, 1] is (0, 1), which is not the empty set.

    Did I understand correctly? I am a little foggy on the properties of real numbers, so I can't really back up my claims about the density of rationals at the moment.

    Thanks for your help!
     
  5. Mar 4, 2005 #4
    yes....you've got it right...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?