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Let $E$ be a set and $(x_n)$ be a sequence of $E$. Suppose that $\lim_{n\to\infty}(x_n)=x$ and that $x$ is an isolated point of $E$. Show that there exists $N\in\mathbb{N}$ so that $x_n=x$ for $n\ge N$.

Here is what I was thinking:

$x$ is an isolated point, so there exists a $c > 0$ such that $(x-c,x+c)\cap E={x}$

Now, we are given $\lim_{n\to\infty}(x_n)=x$.

With $c > 0$, there exists an $N>0$ such that $|x_n-x| whenever $n\ge N$

Now, $|x_n-x| implies $-c or $x-c

Hence, $x_n\in(x-c,x+c)$.

Since $(x-c, x+c)\cap E={x}$, and $x_n\in E$. Thus $x_n=x$ for $n\ge N$. QED

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    Thank you, I will fix that.2012-11-08

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Your proof is correct provided $E \subseteq \mathbb{R}$. If $E$ is a subset of an arbitrary metric space, you would make the same argument but with the open ball $B_c(x)$, and technically $(x - c, x + c) \cap E = \{x\}$