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Confusion about a theorem

  1. Jun 25, 2008 #1
    From page 45 of "Mathematical Analysis" by Tom Apostol:

    3-17 Theorem. If S is closed, then the complement of S (relative to any open set containing S) is open. If S is open, then the complement of S (relative to any closed set containing S) is closed.

    Proof. Assume [tex]S\subset A[/tex]. Then [tex]A-S=E_1-[S\cup(E_1-A)][/tex]. (The reader should verify this equation.) If S is closed and A is open, then [tex]E_1-A[/tex] is closed, [tex]S\cup(E_1-A)[/tex] is closed, [tex]A-S[/tex] is open. The converse is similarly proved.

    Now I'll prove the part that Apostol leaves to the reader:

    Given two subsets A and S of [tex]E_1[/tex], [tex]A-S=E_1-[S\cup (E_1-A)][/tex].

    Proof: If [tex]x\in(A-S)[/tex], [tex]x\in A[/tex] and [tex]x\notin S[/tex]. Thus [tex]x\notin[S\cup(E_1-A)][/tex]. So [tex]x\in E_1-[S\cup (E_1-A)][/tex]. This proves that [tex]A-S\subset E_1-[S\cup(E_1-A)][/tex].

    If [tex]x\in E_1-[S\cup(E_1-A)][/tex], [tex]x\in E_1[/tex] and [tex]x\notin [S\cup(E_1-A)][/tex]. Thus [tex]x\notin S[/tex] and [tex]x\notin(E_1-A)[/tex]. But since [tex]x\in E_1[/tex], this last relation implies [tex]x\in A[/tex]. So [tex]x\in(A-S)[/tex].

    I can't see any part of this whole proof that depends on the fact that [tex]S\subset A[/tex]. Am I missing something? If not, why in the world would the author include that hypothesis?
  2. jcsd
  3. Jun 25, 2008 #2


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    I think that, if S is not entirely inside A, the proof still works and you can still show that A - S is open. But A - S is not the complement of S in A so you can't conclude that S is closed in A. In other words, it doesn't prove your theorem.
  4. Jun 25, 2008 #3
    I'm not sure what you mean by "closed in A". My book defines a "closed" subset of [tex]E_1[/tex] as a subset that contains all its accumulation points.
  5. Jun 25, 2008 #4


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    OK, it looks here like they're using the definition: S is closed in A if it's complement in A is open. Is the equivalence of those two definitions proved somewhere?
  6. Jun 25, 2008 #5


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    Also, a set is closed in A if and only if it contains all of its accumulation points that are in A. for example, The set [0, 1) is NOT closed in R but it is closed in (-2,1).
  7. Jun 25, 2008 #6
    Okay just to clarify, the book has defined "closed" to mean what apparently CompuChip would call "closed in [tex]E_1[/tex]". CompuChip: That is not the definition they're using, although they do prove that a set is closed if its complement is open and vice versa.
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