I'm studying for an upcoming departmental exam, and this was one of the questions given on a previous year's exam.
Let $(a_{n})$ and $(b_{n})$ be real, bounded sequences, and let $(c_{n})$ be the sequence created by interleaving the terms of $(a_{n})$ and $(b_{n})$, $$(c_{n}) = (a_{1}, b_{1}, a_{2}, b_{2},\cdots)$$ Prove that $$\limsup(c_{n})\leq \max\{\limsup(a_{n}), \limsup(b_{n})\}.$$ Further, if both $(a_{n})$ and $(b_{n})$ both converge to $\gamma$, then $(c_{n})$ converges to $\gamma$ also.
The second part is pretty straightforward, no problems there. I've tried a couple of different things to solve the first part, mostly playing around with the various definition of $\limsup$ (supremum of subsequential limits, limit of the supremum of the $m$-tail, the "eventually less than $t$" definition, etc), as I am not sure one would work best here. What really confused me was at one point, I thought it should be the opposite, that the inequality should be reversed:
If you let $A, B,$ and $C$ be the set of subsequential limits of $(a_{n}), (b_{n})$, and $(c_{n})$, respectively $(b_{n})$, then clearly, both $A\subseteq C$ and $B\subseteq C$, as both $(a_{n})$ and $(b_{n})$ are already subsequences of $(c_{n})$. This would then imply that $\sup A\leq\sup C$ and $\sup B\leq\sup C$, and hence, that $\max\{\sup A, \sup B\}\leq\sup C$, which is the opposite of what we are to prove.
I'm probably completely wrong with this and missing something painfully obvious, but if anyone can come up with a proof of the original problem and tell me why my "proof" is wrong, I would greatly appreciate it. Bonus points if you can use the first part to imply the second part, which is what I would imagine the test wants.
You have not done anything wrong. You just proved the easy inequality but you are asked to prove that the reverse inequality is also true.
Consider any subsequence of $(c_n)$. Then it includes a subsequence of $(a_n)$ or it includes a subsequence of $(b_n)$. Can you use this to finish the proof?