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Homework Help: Check that a set is closed, and that another is compact

  1. May 27, 2014 #1
    I have a problem with this excercise. Ironically I think I can manage the part that is supposed to be hardest, here is the problem:

    Let [itex](V,||\cdot||)[/itex], be a normed vector-space.

    a), Show that if A is a closed subset of V, and C is a compact subset of V, then [itex]A+C=\{a+c| a \in A, c \in C\}[/itex] is closed.

    b) Show that if both A and C are compact then A+C is compact.

    I think b was ok,so I tried that first: if we have a sequence from A+C: [itex]\{a_n+c_n\}[/itex], then [itex]\{a_n\}[/itex], has a subsequence converging to a, if we look the original sequence but only the indexes from the subsequence converging to a, then of these indexes of the sequence [itex]\{c_n\}[/itex], must have a subsequence converging to c, and since we then have convergence of a subsequence to a+c, we are done?

    I struggle more with a).

    I thought that I could show that if ther is a sequence from A+C converging to a point, then this point must be in A+C. So I start with the sequence [itex]\{a_n+c_n\}[/itex], which I assume converges to b, I must show that b is in A+C.
    I get that since C is compact there must be a subsequence so that [itex]\{c_{n_k}\}[/itex], converges to an element c in C. Then using the same indexes [itex]\{a_{n_k}\}[/itex] must converge to b-c. But how do I proceed to show that b is in A+C?
    Last edited: May 27, 2014
  2. jcsd
  3. May 27, 2014 #2
    Will the sequence ##(a_{n_k} + c_{n_k})## converge?
  4. May 27, 2014 #3
    Yeah, that is a subsequence of the original sequence. And I started with the original sequence converging, and every subsequence of a converging sequence converges.
  5. May 27, 2014 #4
    So you know that ##(a_{n_k})## converges to ##b-c##. What can you conclude from ##A## being closed?
  6. May 27, 2014 #5
    Yeah I understand, b-c is in A, so (b-c)+c is in A+C.

  7. May 27, 2014 #6
    That's it!
  8. May 27, 2014 #7
    Thanks, vector-spaces are a little tricker than metric-spaces.
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