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Derived set is closed

  1. Jun 9, 2007 #1
    Hi all,

    Would anyone be able to comment if my proof is sound or can be simplified.

    Let (X,d) be a metric space, A a subset of X. The derived set A' of A is the set of all limit points of sequences in A.

    Proposition: A' is closed.

    Proof: Show that its
    complement is open. Let [itex]x\in X\backslash A'[/itex] Then [itex]x\not\in \overline{A}[/itex] so there is a
    non-zero [itex]\varepsilon_x[/itex] s.t. [itex]B(x,\varepsilon_x) \cap A = \emptyset
    \Rightarrow B(x,\varepsilon_x) \subseteq X \backslash A[/itex]. Suppose
    [itex]\exists y \in B(x,\varepsilon_x)[/itex] which in addition belongs to
    A'. Then there's an open ball [itex]B(y,\varepsilon_y) \subseteq
    B(x,\varepsilon_x)[/itex] about y which intersects A. Then
    [itex]B(x,\varepsilon_x) \cap A \neq \emptyset[/itex]. This contradiction shows
    that [itex]B(x,\varepsilon_x) \subseteq X\backslash A'[/itex]
     
  2. jcsd
  3. Jun 9, 2007 #2

    quasar987

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    Please correct me if I'm wrong, but as far as I can see, [tex]A'=\overline{A}[/tex] !
     
  4. Jun 9, 2007 #3
    Now that I think about it, it seems like you are correct. Since surely [itex]A' \supseteq A[/itex]. When I look at my lecture notes/lectures however, the define

    [itex]\overline{A} = A \cup A' [/itex].

    Are there any counterexamples where [itex]A' \neq \overline{A}[/itex]?
     
  5. Jun 9, 2007 #4
    Ahh yes.

    Now that I think about the definition of A', it is not always true that [itex]A'\supseteq A[/itex] since the definition of A' requires that for each x in A there by a point, a ,NOT EQUAL to x s.t. d(x,a) < epsilon for all epsilon > 0.

    [itex]A' \supseteq A[/itex] will fail for any discrete space.
     
  6. Jun 10, 2007 #5

    quasar987

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    I don't understand your last two posts.

    Is it true or not that
    ???

    If so, then we can show that every point in [itex]A'[/itex] is in [tex]\overline{A}[/tex] and vice versa. Firstly, observe that [tex]\overline{A}=A\cup \mbox{Acc}(A)[/tex] (Acc(A) being the set of all accumulation points of A).

    Let x be in A'. If x is in A, then x is in the closure also. If not, then x is obviously an accumulation point of A, so x is in the closure again. This shows that [tex]A'\subset \overline{A}[/tex].

    Now let x be in [tex]\overline{A}[/tex]. Then x is in A or in Acc(A). If x is in A, then define the constant sequence x_n=x. So x is in A'. If x is in Acc(A), then it is also easy to find a sequence of elements of A converging to x, so that x is in A' again. This shows that [tex]\overline{A}\subset A'[/tex].

    So it must be that [tex]A'=\overline{A}[/tex].
     
  7. Jun 12, 2007 #6
    The word limit point and accumulation point are equivalent in the course I'm taking. You cannot claim that x is a limit point because it is the limit of the constant sequence {x,x,...}.

    If you take the discrete space consisting of just two points, say. Then no point in the space is an accumulation point/limit point.

    It is of course true that [itex]A' \subseteq \overline{A}[/itex] since [itex]\overline{A} = A \cup A'[/itex].

    I now think my original post is in error because [itex]x\not\in A' \not\implies x\not\in \overline{A}[/itex].

    Do you have any suggestions on how to correct this?

    thanks

    James
     
  8. Jun 12, 2007 #7

    quasar987

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    To show Acc(A) is closed, show that a sequence of points in Acc(A) lies in Acc(A).
     
  9. Jun 12, 2007 #8
    I'm sure you meant to write:

    To show Acc(A) is closed, show that if a sequence of points in Acc(A) converges to a limit, then the limit lies in Acc(A).
     
  10. Jun 12, 2007 #9

    NateTG

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    The typical example is:
    [tex]A=\{ \frac{1}{n} \}[/tex]
    [tex]A'=\{ 0 \}[/tex]
     
  11. Jun 12, 2007 #10

    quasar987

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    Yes. :smile:
     
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