Set of cluster points of a sequence in $\mathbb{R}$ lies in an interval

Question

If a sequence $\left(a_{n}\right)\in\mathbb{R}$ is bounded, show that $\operatorname{clust}\left(a_{n}\right)\subseteq\left[\inf\left\{S\right\},\sup\left\{S\right\}\right]$, where $S$ is a tail of $\left(a_{n}\right)$ and $\operatorname{clust}\left(a_{n}\right)$ is the set of cluster points of $\left(a_{n}\right)$.

I'm having trouble seeing how this is true. For example, we know that $\inf\left\{a_{n}\right\}\le\inf\left\{a_{m+n}\right\}$ for all $m\in\mathbb{N}$ (here $\left(a_{m+n}:n\in\mathbb{N}\right)$ is the $m$-tail). If a subsequence of $\left(a_{n}\right)$ converged to $\inf\left\{a_{n}\right\}$, then $\inf\left\{a_{n}\right\}\in\operatorname{clust}\left(a_{n}\right)$, but I don't see how we can have that $\inf\left\{a_{n}\right\}\in\left[\inf\left\{S\right\},\sup\left\{S\right\}\right]$.

Answer 1

Indeed, $\inf\left\{a_{n}\right\}\in\left[\inf\left\{S\right\},\sup\left\{S\right\}\right]$ need not be true in general. But in the case that you're considering as an example:

If a subsequence of $(a_n)$ converged to $\inf\{a_n\}$,

it is going to be true. The reason is that if a subsequence of $(a_n)$ converges to $\inf\{a_n\}$, then any tail $S$ of $(a_n)$ contains a tail of that subsequence (simply because a subsequence converging to something is understood to be infinitely long), which still converges to the same limit $\inf\{a_n\}$. Putting together the two facts:

1. that since $S$ is a subset of $(a_n)$, all elements of $S$ are $\ge\inf\{a_n\}$;
2. and that $S$ has a subsequence converging to $\inf\{a_n\}$;

we deduce that $\inf\{a_n\}=\inf(S)$.