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Closed using seq. criterion

  1. Feb 27, 2010 #1
    I'm sorry if this should be in the Analysis forum; I figured it pertained to topology though.
    Let Y be a subspace of a metric space (X,d) and let A be a subset of Y. The proposition includes conditions for A to be open or closed in Y. In class the teacher first proved when A is open and then used complements to prove when A is closed. I'm trying to go the other way around and prove the case when A is closed first, using the sequence criterion. The statement is:
    A is closed in Y iff there is a closed set C in X such that [tex]A=C \cap Y[/tex]
    I haven't really used the sequential criterion for closed sets before, so I don't really know where to start :/ can you provide a starting point? Do I need to construct C or can I abstractly prove the existence of one?
     
  2. jcsd
  3. Feb 27, 2010 #2

    quasar987

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    I'm sure you can at least unwind the definitions and write what it is you need to prove in each of the directions "==>" and "<==".

    This is always a good place to start in a proof.
     
  4. Feb 28, 2010 #3
    Well, for <=
    Suppose [tex]A=C \cap Y[/tex] for C closed in X. Then every convergent sequence in C has its limit in C. Let [tex]x_{n}[/tex] be a convergent sequence in [tex]C \cap Y[/tex]. In particular [tex]x_{n}[/tex] lies in C and hence so does its limit. Then it follows (does it?) [tex]lim_{n \rightarrow \infty} x_{n} \in C \cap Y[/tex]. Hence A is closed in Y.

    For => I'm still not sure, even with the definition.
    Suppose A is closed in Y. Then every convergent sequence in A lies in A. Should I use a contradiction argument and use the fact Y\A is open?
     
  5. Feb 28, 2010 #4

    quasar987

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    You are forgetting an important detail in the definitions.

    A set Y be a subset of a metric space (X,d) and A a subset of Y. Then A is closed in Y if, by definition, for all sequences [itex](y_n)[/itex] with [itex]y_n\in A[/itex] for every n that converges in Y, then that limit is in A.

    What does in mean that a sequence [itex](y_n)[/itex] converges in Y? It means that there exists y in Y such that [itex]\lim_nd(y_n,y)=0[/itex].

    For exemple, take X=R with the usual metric d(x,y)=|x-y|, Y=(0,2), A=(0,1]. Then A is not closed in X since the sequence 1/n converges in X to 0, but 0 is not in A. But, as is easily verified, A is closed in Y. The above argument used above to prove that A is not closed in X does not apply here because 0 is not in Y, so according to the definition above, 1/n does not converge to 0, in Y (!)

    So, with that said, a better formulation of the problem for the direction <== would be:
    "Suppose A=C n Y for C closed in X. Then every sequence in C that converge in X has its limit in C. Let [itex](x_n)[/itex] be a sequence of elements of A that converge in Y, to say, y. We must show that y belongs to A."

    Try solving this, and also rewrite what the hypotheses is and what you want to show for the direction "==>".
     
  6. Feb 28, 2010 #5

    WWGD

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    As a bit of (probably useless) trivia, given a topological space X, and a subspace A

    you can identify limit points (more precisely, points in CL(A); the closure of A) with

    sequences, iff X is first-countable , i.e., if it has a countable local basis. This, of

    course, includes metric spaces. Specifically, in X 1st-countable, we can say that

    a is in CL(A) iff there exists a sequence in A , converging to a. Outside of first-

    countable, we can generalize with nets.
     
  7. Mar 2, 2010 #6
    [tex]x_{n} \in A[/tex] implies [tex]x_{n}[/tex] is in C and Y. Now the limit of the sequence is in Y, so in order to show y is in A, we have to show y is in C. But in particular [tex]x_{n}[/tex] is in X and since C is closed its limit will lie in C, so y is in C, hence is in A, as required.

    Now, for ==> suppose A is closed in Y. That is, if [tex]x_{n} \in A[/tex] for all n is convergent in Y, then its limit is in A. Let C be the set of all sequences in A that converge in X, and its limits. By definition C is closed in X. Then [tex]C \cap Y[/tex] are all sequences in A that converge in Y, and its limit points. But this is just A. Is this correct?
     
  8. Mar 2, 2010 #7

    quasar987

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    Gorgeous!
     
  9. Mar 2, 2010 #8
    Awesome, thanks for the help quasar!
     
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