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Union of intersections

  1. Feb 17, 2009 #1
    Let A, B, and A[tex]\alpha[/tex] denote subsets of a space X.
    neighborhood of [tex]\bigcup[/tex]A[tex]\alpha[/tex] [tex]\supset[/tex] [tex]\bigcup[/tex] neighborhood of A[tex]\alpha[/tex]; give an example where equality fails.

    Criticize the following "proof" of the above statement: if {A[tex]\alpha[/tex]} is a collection of sets in X and if x [tex]\in[/tex] neighborhood of [tex]\bigcup[/tex]A[tex]\alpha[/tex], then every neighborhood U of x intersects [tex]\bigcup[/tex] A[tex]\alpha[/tex]. Thus U must intersect some A[tex]\alpha[/tex], so that x must belong to the closure of some A[tex]\alpha[/tex]. Therfore, x [tex]\in[/tex] [tex]\bigcup[/tex] neighborhood of A[tex]\alpha[/tex].

    I know that the error is in the statement "x must belong to some A[tex]\alpha[/tex] closure over the whole thing." it is false...

    but I don't know how to explain it....

    Can you help me out guys?
  2. jcsd
  3. Feb 17, 2009 #2
    Well, first off, the proof "proves" a reversed inclusion from the one you have listed. Also, I'm not sure what this is looking for: "give an example where equality fails." as you don't have any equality in your problem.

    A few things about the proof though:
    1.) Just because x is in a neighborhood around a point (or a set), this doesn't mean that x is in the closure of that set (i.e., not every neighborhood around x will necessarily intersect the given set). Thus the first statement is incorrect.

    2.) Assuming every neighborhood of x *does* intersect the union, for x to be in the closure of A^a for some a, every neighborhood around x would have to intersect the *same* A^a and it isn't clear that this happens; thus x is not necessarily in the closure of any of the A^a.

    Obviously, the first statement alone is enough to invalidate the entire proof.
  4. Feb 17, 2009 #3
    I don't think the following is false.

    "If x is in the neighborhood of a set A, then x is in the closure of A (cl(A))".

    Let [tex]O \subset A[/tex] be a neighborhood of x. Then O belongs to int(A). Since int(A) belongs to cl(A), O belongs to cl(A). Thus, x is an element of cl(A).

    Please correct me if I am wrong.
  5. Feb 17, 2009 #4
    Ah, but for your proof you have assumed something that we aren't given, namely that x is an element of A. If x is not an element of A, then a neighborhood of x is not a subset of A.

    Also, consider this example: let (X, T) be the discrete topology. Consider any proper subset A of X. Then any point in X\A is also in *a* neighborhood of A, namely X. From this we cannot conclude that every open neighborhood around x intersects A; consider the singleton set {x}. This is an open set in the discrete topology and is thus an open neighborhood around x. But we chose x such that it is not in A. Consequently {x} intersected with A is empty. Thus being in just one neighborhood of a set is not sufficient to be in the closure of that set.

    More generally,

    [tex]x \in \bar{A} \Leftrightarrow \forall \text{neighborhoods U of x}, U \bigcap A \neq \emptyset [/tex]

    Note that we need every neighborhood of x to intersect A, not just one. This was my original point. Sorry if I rambled... :)
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