Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've run into some homework trouble and could use a little help. Here is the question I'm having trouble with:

"Let there be two bounded sequences $\left(a_{n}\right)_{n=1}^{\infty}$ and $ \left(b_{n}\right)_{n=1}^{\infty}$

Show that there exists a strictly monotonically increasing sequence of indexes: $ \left(n_{k}\right)_{k=1}^{\infty} $ in $ \mathbb{N} $ such that both subsequences $\left(a_{n_{k}}\right)_{k=1}^{\infty} $ and $ \left(b_{n_{k}}\right)_{k=1}^{\infty} $ converge."

OK so I know that from the Bolzano–Weierstrass theorem both sequences $\left(a_{n}\right)_{n=1}^{\infty}$ and $ \left(b_{n}\right)_{n=1}^{\infty}$ have some subsequence that converges.

Intuitively I think that the main index sequence should comprise of some kind of combination or union of two different index sequences for two different converging subsequences one for the sequence $\left(a_{n}\right)_{n=1}^{\infty}$ and one for $ \left(b_{n}\right)_{n=1}^{\infty}$

Problem is, I'm stuck, and not sure that my direction is correct (I've tried proving it several times but it fails after I assume something that isn't necessarily correct)

Any hints and help is greatly appreaciated!


share|cite|improve this question
Hint: $(a_n,b_n)$ is bounded. – xavierm02 Dec 6 '13 at 17:35
up vote 6 down vote accepted

Hint: There is a subsequence of the $a_i$ that converges. Look for a subsequence of this subsequence that will deal with the $b_i$ part.

share|cite|improve this answer
Ahh very nice this together with Henry's answer below makes things look really really simple and elegant. Thank you very much! – user475680 Dec 6 '13 at 18:29

Neither a union nor an intersection would work: with a union you might not have convergence, while with an intersection you might not have an infinite subsequence. Instead you could do somethink like:

  • Find an index $n_m$ on which a subsequence of $(a_n)$ converges, say $(a_{n_m})$.

  • Find a subindex $n_{m_j}$ of that first index on which $(b_{n_{m_j}})$ converges.

  • Then the subsubsequence of $(a_{n_{m_j}})$ also converges using the second subindex $n_{m_j}$ with $k=m_j$.

share|cite|improve this answer
Thank you! This is a really nice proof :) I guess I was just over thinking it – user475680 Dec 6 '13 at 18:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.