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I am reading the book Analysis 1 by the author Terence Tao. Let $(a_{n})_{n=m}^{\infty}$ be a sequence of real numbers. Let $a_{N}^{+} = \text{sup}(a_{n})_{n=N}^{\infty}$ and $L^{+} = \limsup_{n\rightarrow\infty} a_{n}$. We need to show that if $L^{+}$ is finite then it is a limit point of $(a_{n})_{n=m}^{\infty}$ (note that $c$ is a limit point of the sequence if $\forall\epsilon > 0\:\forall N\geq m\:\exists n\geq N$ s.t. $|a_{n}-c|\leq\epsilon$). Here is my attempt:

If $L^{+}$ is finite then the monotonically decreasing sequence $(a_{n}^{+})_{n=m}^{\infty}$ is bounded below and therefore we may write $$L^{+} = \lim_{n\rightarrow\infty}a_{n}^{+}$$ Earlier in the book, we have proved the result that if a sequence is convergent then its limit is the unique limit point. So, $L^{+}$ is also a limit point of $(a_{n}^{+})_{n=m}^{\infty}$. So we have $$\forall\epsilon > 0\:\forall N\geq m\:\exists n\geq N\:\text{s.t.}\:|a_{n}^{+}-L^{+}|\leq\epsilon$$ So for a particular choice of $\epsilon$ and $N$ we have $$\exists n\geq N\:\text{s.t.}\:L^{+}-\epsilon\leq a_{n}^{+}\leq L^{+}+\epsilon$$ Now, suppose $L^{+}-\epsilon < a_{n}^{+} = \text{sup}(a_{k})_{k=n}^{\infty}$ then $\exists k\geq n\geq N\:\text{s.t.}\:$ $$L^{+}-\epsilon < a_{k}\leq a_{n}^{+}\leq L^{+}+\epsilon$$ so the claim is proved in this case. However, I do not know how to proceed in the case $L^{+}-\epsilon = a_{n}^{+}$.

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By definition $L^+ = \inf\{a_n^+ : n \geq m\}$ so that $$L^+-\epsilon < L^+ \leq a_n^+$$ for every $n \geq m$.

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  • $\begingroup$ Thank you very much. That was a great observation! $\endgroup$ – Karthik Kannan Jul 17 '19 at 1:00
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Without loss of generality, $m=1.$ For each integer $j$, there is an integer $n_j>j$ such that $a^+_j-1/j<a_{n_j}.$ And of course $a_{n_j}<a^+_j$ so $a^+_j-1/j<a_{n_j}<a^+_j.$ It now follows by the squeeze theorem that $a_{n_j}\to L^+.$

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