Prove that $\limsup a_n = \sup P$ and $\liminf a_n = \inf P$, where $P$ is the set of limit points

Question

Let $\{a_n\}$ be a bounded sequence of real numbers and let $P$ be the set of limit points of $\{a_n\}$. Prove that $\limsup a_n = \sup P$ and $\liminf a_n = \inf P$

---

my work:

Since ${a_n}$ is bounded, then there is an $M$ such that ${a_n} \leq M$ for all $n$. By Bolzano-Weierstrass theorem, ${a_n}$ must have a subsequence that converges (in this case to $P$).
Then, $P \leq M$.
$\liminf {a_n} \leq \limsup {a_n} \leq M$.
Since $P \leq M$, then $\liminf {a_n} \leq \inf P \leq \sup P \leq\limsup {a_n} \leq M$.

I'm not sure how to finish though, or if what I've put is correct

---

Help is greatly appreciated!

Answer 1

This question can be answered by showing that the following two things are true:

1. If $a_{n_k}\to a \in P$, then $\liminf a_n \leq a \leq \limsup a_n$.
2. There is a subsequence that converges to $\liminf a_n$, and one that converges to $\limsup a_n$.

From there, we may conclude that since the liminf and limsup are lower and upper bounds of $P$ that are elements of $P$, they must also be the greatest lower bound and least upper bound.