X contains all its accumulation points $\iff$ X contains all its closure points.

Question

From (say Apostol Analysis) we find the following two facts:

1. X is closed iff it contains ALL its accumulation points.
2. X is closed iff it contains ALL its boundary points.

So, one must conclude that X contains all its accumulation points $\iff$ X contains all its closure points. How to show this fact (without claiming X is closed) ?

Thanks in advance.

Answer 1

$\rightarrow$

Suppose $X$ contains all its accumulation points, then assume that there is a boundary point $x$ which isn't belong to $X$. Hence at every neighbourhood of $x$ there are points of $X$ and of its complement hence $x$ is an accumulation point of $X$, but all accumulation points belong to $X$. Contradiction. Therefore all boundary points belong to $X$

Following that way we can prove the inverse statement.

Answer 2

If $A$ contains all its accumulation points, let $x$ be a boundary point of $A$. If $x \notin A$, then $A$ is an accumulation point of $A$ (as every neighbourhood of $x$ intersects $A \setminus \{x\}$) that's not in $A$ which was not the case. So $x \in A$ and so $A$ contains all its boundary points.

If $A$ contains all its boundary points, let $x$ be an accumulation point of $A$. If $x \notin A$ every neighbourhood of $x$ contains a point in $A$ (by being an accumulation point) and a point not in $A$ ($x$ itself e.g.). So $x$ is boundary point. But $A$ contains all its boundary points, so contradiction and so $A$ contains all its accumulation points.

So both conditions are indeed equivalent.