Prove that a set is closed iff it contains all its accumulation points

Solution 1:

$A\subset X$ is closed $\implies A$ contains all its accumulation points.

Proof by contradiction:

$A\subset X$ is closed $\implies X\setminus A$ is open.

$X\setminus A$ is open $\implies \forall x \in X\setminus A, \exists U_x$ such that $\forall y\in U_x\implies y\in X\setminus A$

Suppose $x$ is an accumulation of $A$ that is not in $A$

$\forall U\in U_x, \exists y\ne x$ with $y \in U\cap A$

$y \in U\cap A \implies y\notin X\setminus A$ -- Contradiction.

$A\subset X$ contains all of its accumulation points $\implies A$ is closed

The other way -- also by contradiction:

Suppose $A\subset X$ contains all of its accumulation points.

Suppose $X\setminus A$ is not open.

There exists an $x \in X\setminus A$ such that $\forall U\in U_x$ there exists $y \in U$ that is also in $A.$

$x$ is an accumulation point, which contradicts the premise.

${\rm QED}$