Let $(A_n), (B_n)$ be two bounded sequences.

Show that there is a sequence of natural numbers $n_1 < n_2 <\cdots$ so that both the subsequences $(A_{n_k})$ and $(B_{n_k})$ converge.

My problem with solving this: Is it possible to say that assuming $A_n > B_n$ for all $n$ then we can make a subsequence of $n_1 < n_2 <\cdots$ from $B_n$ to $\infty$ and then from $A_n$ to $\infty$?

Thanks in advance!

  • 1
    $\begingroup$ First of all, as $(A_n)$ and $(B_n)$ don't relate in any way, it is sufficent, that each bounded sequence has a convergent subsequence. Therefore see en.wikipedia.org/wiki/Bolzano%E2%80%93Weierstrass_theorem $\endgroup$ – k1next Dec 23 '12 at 7:10
  • $\begingroup$ So what does that question mean? I'm puzzled... $\endgroup$ – Harold Dec 23 '12 at 7:28

At this point you are undoubtedly aware of a result stating that a bounded sequence has a convergent subsequence. You should also remember a result stating that a subsequence of a convergent sequence converges. An approach (or extended hints):

  1. Apply the first result to the sequence $A_1,A_2,\ldots$ to get a converging subsequence $A_{n_1},A_{n_2},\ldots$
  2. Apply the first result to the sequence $B_{n_1},B_{n_2},\ldots$ to get - hmm, you think about this a bit
  3. Apply the second result to the sequence $A_{n_{k_1}},A_{n_{k_2}},\ldots$
  • $\begingroup$ Ah Sorry. I thought it was a typo (I thought you meant $B_1,B_2,...$). I am removing my comment $\endgroup$ – Nameless Dec 23 '12 at 7:41
  • $\begingroup$ Actually the first two steps are enough, aren't they? $\endgroup$ – AndreasT Dec 23 '12 at 13:00
  • $\begingroup$ @AndreasT: The third step is admittedly easier, but it is IMHO necessary. $\endgroup$ – Jyrki Lahtonen Dec 23 '12 at 13:28
  • $\begingroup$ @JyrkiLahtonen I'm afraid I misinterpreted 'second result' with 'second step', i.e. I thought you meant to extract again another subsequence. I didn't read carefully enough your answer, I'm sorry... $\endgroup$ – AndreasT Dec 23 '12 at 23:40

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