Let $M$ be a metric space and $A\subseteq M$, and $A'$ is the set of all accumulation points
Prove: $A'$ is closed
Let $a\in M$ and $x_n\in A'$ such that $x_n\to a$, we shall show that $a\in A'$
let $U$ be a neighborhood of $a$, so it has some points of $x_n$, lets take one of them $b=x_{n_0}$ so $U$ is also neighborhood of $b$ so there are infinite points of $A$ and therefore $a\in A'$
We can we conclude that there are infinite point in $U$ and there for $a\in A'$?
Best Answer
Your idea is good, but the exposition should be improved.
You want to prove that every neighborhood of $a$ contains points of $A$ distinct from $a$.
Fix $\varepsilon>0$; by convergence, there is $n$ such that $d(x_n,a)<\varepsilon/2$.
Since $x_n\in A'$ by assumption, there are infinitely many points $b$ of $A$ such that $d(b,x_n)<\varepsilon/2$, in particular a point $b_0$ of $A$ distinct from $x_n$ and $a$ such that $d(b_0,x_n)<\varepsilon/2$.
Then $d(b_0,a)\le d(b_0,x_n)+d(x_n,a)<\varepsilon/2+\varepsilon/2=\varepsilon$.