lim inf, lim sup, limit points

Question

Let $t$ and $s$ be real numbers with $t<s$. Suppose that as $n \to \infty$, $a_{2n} \to t$ and $a_{2n-1}\to s$.

True or False: It must be that $\liminf a_{n} = t$ and $\limsup a_{n} = s$ and $(a_n)$ has no other limit points.

Prove the statement or give a counterexample.

I think true, based on the following example, which I adapted from a wikipedia image (where s=1, t=-1 and $a_{2n-1}$ is the pink sequence and $a_{2n}$ is the light blue sequence):

Hence, all of what the statement said is true... but i don't know how to prove it.

Answer 1

I will use the following definition of $\limsup$ and $\liminf$:

Let $E$ be the set of subsequential limits of $(a_n)$. Then $\limsup_{n \to \infty} a_n := \sup E$ and $\liminf_{n \to \infty} a_n := \inf E$.

It suffices to show that in the setting of your question, $E = \{t, s\}$. Clearly $t,s \in E$. Now suppose $(a_{n_k})$ is a subsequence of $(a_n)$. The sequence of indices $n_k$ must either contain infinitely many even numbers or infinitely many odd numbers (or both).

Edit: This implies that $(a_{n_k})$ contains a further subsequence that either converges to $s$ or to $t$. Thus if $(a_{n_k})$ converges, it must be to either $s$ or $t$. (But it might not converge.)

So $E = \{t,s\}$.

Answer 2

Yes, it is true. Note that $\limsup_na_n$ is the greatest number that can be obtained as the limit of a subsequence of the sequence $(a_n)_{n\in\mathbb N}$. So, let $(a_{n_k})_{k\in\mathbb N}$ be such a convergent subsequence. If infinitely many of the $n_k$'s are even, then $(a_{n_k})_{k\in\mathbb N}$ has a subsequence converging to $t$ and therefore, since we are assuming that the sequence $(a_{n_k})_{k\in\mathbb N}$ converges, its limit can only be $t$. By the same argument, if infinitely many of the $n_k$'s are even, then $\lim_{k\to\infty}a_{n_k}=s$. So, $t$ and $s$ are the only possible limits of subsequences of the sequence $(a_n)_{n\in\mathbb N}$. Since $s>t$, this proves that $\limsup_na_n=s$. By the same argument, $\liminf_na_n=t$.