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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!

Thanks

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Hint: $(a_n,b_n)$ is bounded. –  xavierm02 Dec 6 '13 at 17:35

2 Answers 2

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.

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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 $k=m_j$.

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

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