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Homework Help: Convergent sequences in Cartesian product of vector spaces

  1. Oct 2, 2012 #1
    If A and B are vector spaces over ℝ or ℂ show that a sequence (a_n, b_n) in A×B converges to (a,b) in A×B only if a_n converges to a in A and b_n converges to b in B as n tends to infinity.

    To me this statement sounds pretty intuitive but I have been having trouble actually proving it properly.

    My first attempt was to assume that a_n converges to a in A and b_n converges to b in B then it was kind of easy to see that (a_n, b_n) converges to (a,b) but showing that the converse is true seems to be a bit trickier.

    To me it seems like if you have a sequence (a_n, b_n) where a_n is in A and b_n is in B that it is 'obvious' that it converges to (a,b) only if if the individual sequences converge in their space. I mean like, if it didn't converge in it's space, how could it converge in the Cartesian product?

    Does anyone have any ideas on finishing off this proof? (assuming that I started with the correct idea)
  2. jcsd
  3. Oct 2, 2012 #2
    Vector spaces?? How can you talk about convergence in just vector spaces?
  4. Oct 2, 2012 #3
    Do you mean that it should be a normed vector space?

    Like in this case |(a,b)| = |a|_A + |b|_B for a in A and b in B

  5. Oct 2, 2012 #4
    Yes, A and B should be normed vector spaces (or metric spaces more generally). And you should make a norm on AxB as well. Your proposal of the norm


    is a good one.

    Now, can you write out what it means that the sequence [itex](a_n,b_n)[/itex] converges in AxB? And what it means that [itex]a_n[/itex] converges in A and that [itex]b_n[/itex] converges in B?
  6. Oct 2, 2012 #5
    For the cartesian product,

    For some ε>0 there exists an N such that |(a_n,b_n) - (a,b)|<ε whenever n>N

    |(a_n-a,b_n-b)| = |a_n-a|_A + |b_n-b|_B

    For individual spaces

    For some ε>0 there exists an N such that |a_n - a|_A<ε/2 whenever n>N
    (same definition for b_n)

    and if you add them together you will get

    |a_n - a|_A + |b_n - b|_A<ε/2+ε/2=ε

    but then to prove it don't you also need to show that the converse of the statement is not true?

    Is it enough to say that

    For some ε>0 there exists an N such that |(a_n,b_n) - (a,b)|<ε whenever n>N

    |(a_n-a,b_n-b)| = |a_n-a|_A + |b_n-b|_B<ε

    but if there does not exist an N such that |a_n-a|_A <ε/2 for n>N then it doesn't converge ?
  7. Oct 2, 2012 #6
    Same for the [itex]b_n[/itex], but the N there might be different than the N for the [itex]a_n[/itex]. So how do you solve that?

    What do you mean?
  8. Oct 2, 2012 #7
    Sorry I editing my post a bit to make it make more sense haha,

    If the N are different then for the a case i'll write it as N_a, then you need N = max(N_a, N_b) right?
  9. Oct 2, 2012 #8
    Are you trying to show now that if [itex](a_n,b_n)[/itex] converges, then [itex]a_n[/itex] converges. To prove this you have to find for all [itex]\varepsilon>0[/itex] an N such that for n>N holds that [itex]|a_n-a|_A<\varepsilon[/itex]. Ho< would you find that N?

  10. Oct 2, 2012 #9
    Wouldnt that N be the same N required for (a_n,b_n) to converge?
  11. Oct 2, 2012 #10
    Yes, try to use that N. So you know that


    and you must prove that


    Can you do this?
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