# Doubt in a part of the proof of Bolzano-Weierstraß Theorem

Let $$n$$ be a natural number and $$\{a_n\}$$ be a bounded sequence of real numbers, that is $$\vert a_n\vert\leq M$$, for all $$n$$ ($$M\geq0$$). Define $$E_n=\overline{\{a_j\vert j\geq n\}}$$ as the closure of the set $$\{a_j\vert j\geq n\}$$. Then $$E_n\subseteq[-M, M]$$

As far as I know, the closure of a set can be defined as the set itself plus all its boundary points. Henceforth, since a boundary point can either belong or not to the set it bounds, I would say that $$[-M, M]\subseteq E_n$$ and NOT that $$E_n\subseteq[-M, M]$$.
A counterexample to $$E_n\subseteq[-M, M]$$ could be represented by a boundary point of the set $$\{a_j\vert j\geq n\}$$ which is not in the interval $$[-M, M]$$, that is which does not belong to the set it bounds.

Could you please clarify such a doubt and highlight the flaws of my reasoning?

• "I would say that $[-M, M]\subseteq E_n$". Why? – Angina Seng May 31 at 10:23
• Since..if my definition of closure set is correct, if I am in the interval $[-M, M]$ I am for sure in the closure set of the set $\{a_j\vert j\geq n\}$ (which is bounded - above and below - by $M$) @AnginaSeng – Strictly_increasing May 31 at 10:27
• So, if you have a sequence $a_n=1/n$ which is bounded by $M=1$, then every number between $-1$ and $1$ is in the closure of the set $\{1/1,1/2,1/3,\ldots\}$??? – Angina Seng May 31 at 10:30
• Which is the correct definition of closure set? Could you please give me a reference to it? Since I guess that the one I have provided is wrong – Strictly_increasing May 31 at 10:34
• Are you aware that $\{a_j \mid j \geq n\} \subset [-M,M]$ and that, since $[-M,M]$ is closed, we have $[-M,M] = \overline{[-M,M]}$? – h3fr43nd May 31 at 13:37

Recall $$A \subseteq B \implies \overline{A} \subseteq \overline{B}$$ and $$B = \overline{B}$$ iff $$B$$ is closed.
Now, apply this with $$A= \{a_j: j \geq n\}$$ and $$B = [-M,M]$$.