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Let $(x_n)$ a bounded sequence of real number and let $E$ the set of accumulation point. Prove that $$\sup E=\limsup_{n\to \infty }x_n.$$


Attempts

I set $\displaystyle\lambda =\limsup_{n\to \infty }x_n$ and $\beta =\sup E$. I proved that $E\neq \emptyset$ and $E$ bounded, so $\beta $ is well defined. Then I proved that $\lambda \in E$ and thus $\lambda \leq \beta $. I'm now trying to prove that $\beta \leq \lambda $ but I really have problem for this. I know that $\beta $ is an adherence point of $E$, and thus there is a sequence $(\beta _m)_m$ that converge to $\beta $. Let $m\in\mathbb N$. Since $\beta _m$ in $E$, there is a subsequence $(x_{n_k})$ of $(x_n)$ that converge to $\beta _m$.

How from this I can construct a subsequence of $(x_n)$ that converge to $\beta $ ?

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  • $\begingroup$ I would prove that $\beta > \lambda$ leads to a contradiction. Namely, if that is true, $\lambda$ cannot be $\lim \sup x_n$ $\endgroup$ – Joe May 8 '18 at 15:05
  • $\begingroup$ When you have a sequence of sequences $x_{n}^{(m)}$ such that their limits $x_n^{(m)}\to \beta_m$ converge $\beta_m\to \beta$, then you can construct a new sequence that converges to the limit, by taking the diagonal sequence $x_n^{(n)}\to \beta$. $\endgroup$ – user551819 May 8 '18 at 15:23
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Hint

You don't need to construct a subsequence of $(x_n)$ that converge to $\lambda $. Set $\displaystyle y_n=\sup_{k\geq n}x_k.$ We have that $y_{n_k}\geq x_{n_k}$ for all $k$. Letting $k\to \infty $ you get that $\beta _m\leq \lambda $ for all $m$. The claim follow.

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  • $\begingroup$ Thanks a lot @Surb. I see now. $\endgroup$ – user352653 May 8 '18 at 15:23

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