TITLE: Determine set of subsequential limits
QUESTION [5 upvotes]: Suppose you are given a sequence $\{s_n\}$, where the odd terms $\{s_{2m+1}\}$ and the even terms $\{s_{2m}\}$ form two converging subsequences: $$s_{2m+1} \to \alpha,$$ $$s_{2m} \to \beta.$$
My intuition is that the set of subsequential limits $E$ of $\{s_n\}$ equals $\{\alpha, \beta\}$, but I’m struggling with a proof. Is there a way to pick another subsequence of $\{s_n\}$ that converges to some $\gamma \neq \alpha, \gamma \neq \beta$?
@David
If a subsequence contains finitely many odd or finitely many even terms, then in the tail it converges to either $\alpha$ or $\beta$. If a subsequence contains infinitely many odd terms and infinitely many even terms, then for no $N$ does the tail $\{s_n: n \geq N\}$ get arbitrarily close to either $\alpha$ or $\beta$. Is that the right reasoning?

REPLY [2 votes]: You have the right idea. 
(I). In $\Bbb R,$ one useful way of stating that $\lim_{n\to \infty}x_n=x$ is that for every $r>0$ the set $\{n: x_n\not \in (-r+x,r+x)\}$ is finite.
(II). In the Q, if $\alpha \ne \gamma \ne \beta$ then there exists $r>0$ such that $$(-r+\gamma,r+\gamma)\cap \;[(-r+\alpha,r+\alpha)\cup (-r+\beta,r+\beta)]=\phi.$$ For example $r=\frac {1}{2} \min (|\alpha -\gamma|, |\beta -\gamma|).$ 
For such $r$ we have $$\{n: s_n  \in (-r+\gamma, r+\gamma)\}\subset A \cup B$$ where $A=\{2m+1: s_{2m+1}\not \in (-r+\alpha,r+\alpha)\}$ and $B=\{2m: s_{2m}\not \in (-r+\beta,r+\beta)\}.$ 
By (I), $A$ and $B$ are finite, so $A\cup B$ is finite, so $\{n:s_n\in (-r+\gamma,r+\gamma)\}$ is finite. But if a subsequence of $(s_n)_n$ converged to $\gamma$ then there would be infinitely many $n$ such that $s_n\in (-r+\gamma,r+\gamma).$