# Why the converse to Rudin 3.2(d) is not generally true?

Rudin 3.2(d): If $X$ is a metric space and $E$ is a subset of $X$ and $p$ is a limit point of $E$, then there is a sequence $\{p_n\}$ in $E$ such that $\{p_n\}$ converges to $p$.

Why the converse: if there is a sequence converging to $p$, then $p$ is a limit point, is not generally true?

• It's definitional. Some authors will say that every element of $E$ is a limit point of $E$, precisely because of the constant sequence. To answer your question precisely, we need to know which definition of limit point you're using. – dbx Nov 20 '17 at 17:23

Let p be a non limit point. Thus the sequence $\{x_n\}_{n\in\mathbb N},\ x_n=p\forall n\in \mathbb N$ converges to p.
• @ZhouChenliang No, you'd need "an infinite number of elements of the sequence are no equal to $p$." Or, equivalently, the sequence is not ultimately equal to $p$ identically. – Clement C. Nov 20 '17 at 17:27
• @ZhouChenliang The quantifiers/phrasing are confusing. Did you mean "not(every term of the sequence is equal to $p$)" (not every term of the sequence is equal to $p$, as I read it), or "every term of the sequence is not equal to $p$" ("no term of the sequence is equal to $p$")? – Clement C. Nov 20 '17 at 17:34
If the sequence $(p_{n})$, $p_{n}\ne p$, given $\epsilon>0$, choose some $N$ such that $p_{n}\in B_{p}(\epsilon)$ for all $n\geq N$. So $p_{n}\in B_{p}(\epsilon)-\{p\}$ for all such $n$, the neighbourhood even contains infinitely many $p_{n}$, so it is a limit point.
As long as the sequence does not eventually equal to $p$, then it goes through.