1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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

    Or?
     
  5. Oct 2, 2012 #4

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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

    [tex]\|(a,b)\|=\|a\|_A+\|b\|_B[/tex]

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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?

    Correct.
     
  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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Yes, try to use that N. So you know that

    [tex]\|(a_n,b_n)-(a,b)\|=\|a_n-a\|_A+\|b_n-b\|_B<\varepsilon[/tex]

    and you must prove that

    [tex]\|a_n-a\|_A<\varepsilon[/tex]

    Can you do this?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Convergent sequences in Cartesian product of vector spaces
Loading...