-1
$\begingroup$

Suppose $( a_n ) _n$ and $( b_n ) _n$ are two sequences of real numbers (not necessarily Cauchy or convergent) Suppose $| a_n | < 2 \ \forall n$ and $| b_n | < 17 \ \forall n$. Prove that there is a common convergent subsequence; that is, there exists a subsequence $(m_k)$ so that $( a_{m_k} )$ and $( b_{m_k} )$ are both convergent.

I know that there exists $( n_k )$ so that $( a_{n_k} )$ is convergent because every bounded sequence in the reals has a convergent subsequence.

Can I say that the sequence $( b_{n_k})$ is bounded?

$\endgroup$
  • $\begingroup$ Yes: every subsequence of $\langle b_n:n\in\Bbb N\rangle$ is bounded, since the original sequence is. $\endgroup$ – Brian M. Scott Nov 8 '15 at 21:04
0
$\begingroup$

Yes, your idea is correct. Since $( b_n ) _n$ is bounded, then $( b _{n_k} ) _k$ is bounded too, so it too admits a convergent sub-sequence $(b _{n _{k_i}}) _i$. Now, since the sub-sequence $( a _{n_k} ) _k$ was convergent, so will be its sub-sequence $(a _{n _{k_i}}) _i$. Therefore, the sub-sub-sequences $(a _{n _{k_i}}) _i$ and $(b _{n _{k_i}}) _i$ are simultaneously convergent (not necessarily to the same limit, of course).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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