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|
Now, $|x_n-x|
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