I have been trying to prove this exercise for way too long and am nowhere near a proof. I have asked this question before and haven't got a satisfactory answer. This could be my own fault for not defining everything properly.
The book I am learning from(Analysis $1$ by Tao) defines the limit superior and limit inferior as follows. Some people like using the notation used in this wikipedia article. The wikipedia article is exactly the same as the following definitions.
The following is how the book defines limit superior and limit inferior.
Sequence of supremums: Let $(a_n)_{n=1}^\infty$ be a sequence of real numbers. Define a new sequence $(a^+_N)_{N=m}^\infty$ by $a^+_k:=\sup ((a_n)_{n=k}^\infty)$.
Limit superior: The limit superior of a sequence $(a_n)_{n=m}^\infty$ is defined to be $\inf ((a^+_N)_{N=m}^\infty)$
Sequence of infimums: Let $(a_n)_{n=m}^\infty$ be a sequence of real numbers. Define a new sequence $(a^-_N)_{N=m}^\infty$ by $a^-_k:=\inf ((a_n)_{n=k}^\infty)$.
Limit inferior: The limit inferior of a sequence $(a_n)_{n=m}^\infty$ is defined to be $\sup((a^-_N)_{N=m}^\infty)$.
Limit point: Let $(a_n)_{n=m}^\infty$ be a sequence of real numbers. Let $c$ be a real number. Then $c$ is said to be a limit point of $(a_n)_{n=m}^\infty$ if for every $\varepsilon>0$ and for every $N\ge m$, there exists an $n\ge N$ such that $|a_n-c|<\varepsilon$.
The exercise asks to prove that if $c$ is a limit point of a sequence of real numbers $(a_n)_{n=m}^\infty$. And $L^-$ and $L^+$ are the limit inferior and limit superior respectively, then $L^-\le c \le L^+$.
I have previously been suggested proofs using subsequences. The only problem is that the book hasn't introduced subsequences yet. Which means that the proof shouldn't involve subsequences. Could someone guide me on how to prove this statement?
Best Answer
Suppose $A \le a_n \le B$ for a infinitely many $n$, then we must have $\liminf_n a_n \le B$ and $\limsup_n a_n \ge A$.
If $c$ is a limit point of $a_n$, pick some $\epsilon >0$ then $c-\epsilon < a_n < c+ \epsilon$ infinitely often. Hence $\liminf_n a_n \le c+ \epsilon$ and similarly $\limsup a_n \ge c-\epsilon$.
Since this is true for any $\epsilon>0$, the result follows.