Let $A$ be a set , in a metric-space $(X,d)$ , having no limit points ; then I'm trying to prove that every convergent sequence $(x_n)$ in $A$ is ultimately constant . Please see the proof and tell whether its right or not
Proof:- Since $(x_n)$ is convergent , it has a unique limit say $x$.We will assume that $(x_n)$ is not ultimately constant and show that $A$ must have a limit point. If $\exists k \in \mathbb N$ such that $x_n=x , \forall n \ge k$ then $(x_n)$ is ultimately constant ; so assuming $(x_n)$ is not ultimately constant , $\forall n \in \mathbb N , \exists r_n \ge n$ such that $x_{r_n}\ne x$ . Consider any $\epsilon>0$ , then $\exists m \in \mathbb N$ such that $d(x_n ,x)< \epsilon , \forall n \ge m$ , so $\exists r_m \ge m$ such that $x_{r_m}\ne x$ and we also have $d(x_{r_m},x)<\epsilon$ i.e. $x_{r_m} \in B(x,\epsilon)\cap A $ and $x_{r_m} \ne x$ , hence $ \Big(B(x,\epsilon)-\{x\}\Big)\cap A \ne \phi $ , since $\epsilon>0$ is arbitrary , we conclude $x$ is a limit point of $A$.
Is the proof ok ?
Your proof is correct. But choosing the "contrapositive route" you have unnecessarily complicated matters.
Assume $\lim_{n\to\infty} x_n=x\in A$. Since $x$ is not a limit point of $A$ there is an $\epsilon>0$ with $B(x,\epsilon)=\{x\}$. Now there is an $n_0$ with $x_n\in B(x,\epsilon)$ for all $n>n_0$. It follows that $x_N=x$ for all $n>n_0$.