I understand that similar questions to this have been asked here (1) and here (2) but I believe my question is different because with regards to (1) I am not asking about the set $X$ that contains the points of my sequence and with regard to (2) because I have the distinct definitions of limit and limit point in mind. That said:
My Question: My textbook, Fundamental Ideas of Analysis by Michael Reed, defines a limit point as:
Definition: Let $(a_n)$ be a sequence of real numbers. A number $d$ is called a limit point if, given $\epsilon > 0$ and an integer $N$, there exists $\color{red}{\mathrm{ an }} \ n \ge N$ so that $|a_n – d| \le \epsilon$.
The definition of a limit is as follows:
Definition: We say that a sequence $(a_n)$ converges to a limit $a$ if, for every $\epsilon \ge 0$, there is an integer $N(\epsilon)$ so that $|a_n – a| \le \epsilon \color{red}{\text{ for all }} n \ge N$.
From these definitions, and given this question:
If a sequence converges it has exactly one limit point. Is the converse true? If not, provide a counterexample
Can I conclude that the converse is indeed true simply by appealing to the definitions, or do I need a tighter proof. And if I do, I'm a little stuck on how to refine this proof.
Best Answer
The sequence $2,1/2,3,1/3,4,1/4 \cdots, n,1/n \cdots$ has the only limit point $0$, but not converges.