Equivalence of Definitions of Closed Set

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Let $H \subseteq S$.

Proof
Let $T = \left({S, \tau}\right)$ be a topological space.

Let $H \subseteq S$.

Definition 1 implies Definition 2
Let $H$ be a closed set of $T$ by definition 1.

Let $H^{\complement}$ denote the relative complement of $H$ in $S$.

By definition of closed set in $T$:
 * $H^{\complement}$ is open in $T$

From Set is Open iff Neighborhood of all its Points:
 * $\forall x \in S: x \notin H \implies H^{\complement}$ is a neighborhood of $x$.

By definition of limit point:
 * $\forall x \in S: x \notin H \implies x$ is not a limit point of $H$

Thus $H$ is a closed set of $T$ by definition 2.

Definition 2 implies Definition 1
Let $H$ be a closed set of $T$ by definition 2.

Then by definition: $\forall x \in S: x \notin H \implies x$ is not a limit point of $H$

By definition of limit point: $\forall x \in S: x \notin H \implies H^{\complement}$ is a neighborhood of $x$

By Set is Open iff Neighborhood of all its Points: $H^{\complement}$ is open in $T$

Thus $H$ is a closed set of $T$ by definition 1.