I am new to Real Analysis, and I have found this problem hard to formalize.
Problem
Let $(p_n)_{n\in\mathbb{N}}$ and $(q_n)_{n\in\mathbb{N}}$ sequences such that $(p_n)\to u$ and $(q_n)\to v$. Consider the sequence $(x_n)_{n\in\mathbb{N}}$ such that $x_n:=\begin{cases}q_m\text{ if } n = 2m-1\\ p_m\text{ if } n = 2m\end{cases}$. Show that $\limsup_{n\to\infty}x_n=\max\{u, v\}$, and $\liminf_{n\to\infty}x_n=\min\{u, v\}$.
My attempt
I begin dividing into two cases. Either $u = v$, or $u\neq v$. For the first case, it seems easy to conclude that since every subsequence converges to $u=v$, then $\lim_{n\to\infty}x_n=\limsup_{n\to\infty}x_n=\liminf_{n\to\infty}x_n=u=v$.
For the second case, without loss of generality let $u>v$. We want to show then, $\limsup_{n\to\infty}=u$ and $\liminf_{n\to\infty}=v$.
Let us start with the first aim (as the limit inferior could be proved analogously):
Since $u>v$, $\exists\epsilon_1>0: u > v + \epsilon_1.\tag{1}$
Since $q_{m}\to v$ and $p_{m}\to u$, for all $\epsilon>0$, $\exists M_q$ s.t. $m>M_q, |x_{2m-1}-v|<\epsilon$, and $\exists M_p$ s.t. $m>M_p, |x_{2m}-u|<\epsilon$.
Let $\epsilon>0$. Set $M=\max\{M_q, M_p\}$. Then, for $m>M$, $|x_{2m-1}-v|<\epsilon$ and $|x_{2m}-u|<\epsilon$, i.e.: $$v-\epsilon<x_{2m-1}<v+\epsilon\quad\text{ and }\quad u-\epsilon<x_{2m}<u+\epsilon.$$
If $\epsilon<\frac{\epsilon_1}{2}$, $$v-\epsilon<x_{2m-1}<v+\epsilon<v+\frac{\epsilon_1}{2}\quad\text{ and }\quad u-\frac{\epsilon_1}{2}<u-\epsilon<x_{2m}<u+\epsilon,$$
and because of (1), $$v-\epsilon<x_{2m-1}<v+\epsilon<v+\frac{\epsilon_1}{2}<u-\frac{\epsilon_1}{2}<u-\epsilon<x_{2m}<u+\epsilon,$$ that is, $x_{2m-1}<u-\epsilon<x_{2m}<u+\epsilon$ for all $m>M$.
Let $i>2M$ and $s_i=\sup(\{x_k\}_{k\geq i})$, then $s_i>u-\epsilon$ because $u-\epsilon$ is not an upper bound of $\{x_k\}_{k\geq i}$. On the other hand, $s_i\leq u+\epsilon$, since it is an upper bound. Therefore,
for $0<\epsilon<\frac{\epsilon_1}{2}$ and $i>2M$, $$u-\epsilon < s_i \leq u+\epsilon\tag{2}\Rightarrow|s_i-u|\leq\epsilon.$$ Now, $\limsup_{n\to\infty}x_n=u$ if the inequality in (2) is strict, but I cannot see how to overcome this issue.
And the other concern is: what happens when $\epsilon>\epsilon_1/2$? My intuition is that we can set the same $M$ as for any $\epsilon^{\ast}<\epsilon_1/2$, and that ensures $|s_i-u|<\epsilon^{\ast}<\epsilon$ for any $\epsilon>0$. But how can I state this fact rigorously?
I feel this proof can probably be shortened and/or better written, so any improvement or advice is more than welcome.
Best Answer
Your proof is good, sorry I did not read it carefully earlier. I fell in love with the subsequence proof, but adding in the subsequence characterization made it pretty long.
To address the issue with $|s_i-u|\leq \epsilon$ in equation (2). Between “Let $\epsilon>0$” and up to an including equation (2), replace $\epsilon$ with $\epsilon/2.$ Then $|s_i-u|\leq \epsilon$ becomes $|s_i-u|\leq \epsilon/2<\epsilon$. If you end up with $\,$ but need $<$, this trick almost always works.
The other issue you mention is what happens when $\epsilon\geq\epsilon_1/2$ (now $\epsilon/2\geq\epsilon_1/2.$) In this case the $M$ you found to work for $\epsilon=\epsilon_1/4$ will work. This type of argument is the reason you always can assume $\epsilon$ is small, at least smaller than any fixed number $>0$.
Let me know if you need more clarification.