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?

  • $\begingroup$ 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

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).


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