1
$\begingroup$

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|$$-c or $$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

  • 0
    Your proof looks good.2012-11-08
  • 0
    Your argument looks fine but your statement $$\text{x is an isolated point, so by definition, $(x−c,x+c)\cap E=x$}$$ is incorrect. The definition of an isolated point goes as follows $$\text{$x$ is an isolated point, if there exists a $c>0$ such that $(x-c,x+c) \cap E = x$}$$ Hence, if $x$ is an isolated point, then ***there exists a $c>0$*** such that $(x-c,x+c) \cap E = x$.2012-11-08
  • 0
    Thank you, I will fix that.2012-11-08

1 Answers 1