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