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Let $A\subseteq M$ metric space and $A'$ be the set of the limit points of $A$

Prove: $A'$ Is Closed

Let $a\in M$ and $x_n\to a$ such that $x_n\in A'$ we will show that $a\in A'$

let $U$ be a neighborhood of $a$ so $U$ contains points from $\{x_n\}\subseteq A'$ if we will take $b=x_{n_0}$ so $U$ is also a neighborhood of $b$ and therefore contains infinite many points of $A$ therefore $a\in A'$

why can we conclude that $a\in A'$?

newhere
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1 Answers1

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$x_n \in A^\prime$ and so you can find a $y_n \in A$ with $d(x_n,y_n) <1/n$. Now you can prove that the sequence $y_n$ converges to $a$ (this is because $d(y_n,a) \leq d(y_n,x_n)+d(x_n,a) \leq 1/n + d(x_n,a) \to 0$ as $n\to\infty$) and so we know that $a$ is a limit point ($a\in A^\prime$) and thus $A^\prime$ is closed.

Jack
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