Does $s_n$ bounded but not convergent implies at least two different convergent subsequences? If a sequence of real numbers $(s_n)$ is bounded but not convergent, does this imply that there are at least two subsequences of $(s_n)$ that converge to different numbers? I know if a sequence is bounded, then it has a convergent subsequence but I am not entirely sure if my guess is valid.
 A: What you are looking to prove is true. This is because a bounded sequence is convergent if and only if it has exactly one accumulation point. Here, a real number $x$ is said to be an accumulation point of a sequence $(x_n)$ if there is a subsequence $(x_{n_k})$ converging to $x$.
For a more direct proof, let $(x_n)$ be bounded sequence that does not converge and let $(x_{n_k})$ be a subsequence converging to some $x \in \mathbb{R}$ (such a sequence exists by compactness). Because $x_n$ does not converge to $x$, there exists $\varepsilon_0 > 0$ and a corresponding subsequence $(x_{n_j})$ such that
$$
|x_{n_j} - x| \geq \varepsilon_0, \quad \forall j \geq 1.
$$
Now, this subsequence $(x_{n_j})$ must also bounded and hence has a subsequence that converges to some $y \in \mathbb{R}$. But, by the above, $y$ cannot be $x$. Hence, we have found two subsequences converging to different points.
Note that it is critical here that the sequence $(x_n)$ be bounded. It's also easy to come up with an example of an unbounded sequence for which this fails. For instance, the sequence
$$
x_n := \begin{cases}
n & \text{if $n$ is odd},\\
1 & \text{if $n$ is even}
\end{cases}
$$
does not converge but has only a single accumulation point.
A: Adding to the answer here: $\lim \inf s_n$ and $\lim \sup s_n$ would be explicit examples of accumulation points of a bounded sequence of real numbers. The convergent sequences are exactly those for which these two quantities are equal.
