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Int(A) + ext(A) not dense

  1. Jan 8, 2008 #1
    Can someone help me find an example of how the union of int(A) and ext(A) doesn't have to be dense in some space X? Thanks.
  2. jcsd
  3. Jan 8, 2008 #2


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    Define your terms. What are int(A) and ext(A)? The former I'm assuming is the interior of A, and the latter is ... the interior of the complement of A (= the complement of the closure of A)? And post your thoughts on the matter.
  4. Jan 9, 2008 #3
    int(A) = interior of A
    ext(A) = exterior of A, or the interior of the complement of A

    My thoughts are thus: int(A) and ext(A) are both open sets, so their union is an open set, and if we let B = union of int(A) and ext(A) then B = int(B). So the only way the closer is not equal to the entire space (making it dense) would be if the complement of B had some open points so that ext(B) was not empty.

    However, since the complement of B is a subset of the complement of A, and ext(A) is all the open points of of the complement of A, the only way I can see that this would have a chance of being possible is if somehow you could construct a space where the int(A)=ext(A) for some set in that space (neither of which equaled the entire space). But I can't figure out how to construct a space where this is possible.

    Those are my thoughts.
  5. Jan 9, 2008 #4


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    What in the world are "open points"? Do you mean "interior points" of a given set? In general topology, points do not have any properties- "points are points".
  6. Jan 9, 2008 #5
    Sorry, bad phrasing. I figured it should have been obvious what I meant from context. I believe my book calls them "interior points", which are points that are contained in some open set that is completely contained in the set in question.
  7. Jan 9, 2008 #6
    Nevermind, I got it. I forgot about the possibility that int(A) and ext(A) could both be the empty set, and thus the closure would also be the empty set (which I guess meets my criteria anyway of int(A) = ext(A)).
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