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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?

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  • $\begingroup$ "I would say that $[-M, M]\subseteq E_n$". Why? $\endgroup$ – Angina Seng May 31 at 10:23
  • $\begingroup$ 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 $\endgroup$ – Strictly_increasing May 31 at 10:27
  • $\begingroup$ 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 $\{1/1,1/2,1/3,\ldots\}$??? $\endgroup$ – Angina Seng May 31 at 10:30
  • $\begingroup$ 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 $\endgroup$ – Strictly_increasing May 31 at 10:34
  • $\begingroup$ Are you aware that $\{a_j \mid j \geq n\} \subset [-M,M]$ and that, since $[-M,M]$ is closed, we have $[-M,M] = \overline{[-M,M]}$? $\endgroup$ – h3fr43nd May 31 at 13:37
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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 $B = [-M,M]$.

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