# Showing that the limit superior is a limit point of a sequence

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 $$L^{+}-\epsilon = a_{n}^{+}$$.

By definition $$L^+ = \inf\{a_n^+ : n \geq m\}$$ so that $$L^+-\epsilon < L^+ \leq a_n^+$$ for every $$n \geq m$$.
Without loss of generality, $$m=1.$$ For each integer $$j$$, there is an integer $$n_j>j$$ such that $$a^+_j-1/j And of course $$a_{n_j} so $$a^+_j-1/j It now follows by the squeeze theorem that $$a_{n_j}\to L^+.$$