Equivalence of Definitions of Closure of Topological Subspace

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

Let $H \subseteq S$.

The following definitions of topological closure of $H$ in $T$ are equivalent:

Definition 5
Let $H^-$ denote the closure of $H$.

Then:


 * $(1): \quad H^- = \displaystyle \bigcap \left\{{K \supseteq H: K}\right.$ is closed in $\left.{T}\right\}$
 * $(2): \quad H^-$ is the smallest closed set that is a superset of $H$
 * $(3): \quad H^-$ is the union of $H$ and the boundary of $H$
 * $(4): \quad H^-$ is the union of all isolated points of $H$ and all limit points of $H$

Definition $(1)$ is equivalent to Definition $(2)$
This is proved in Set Closure as Intersection of Closed Sets.

Definition $(2)$ is equivalent to Definition $(3)$
This is proved in Set Closure is Smallest Closed Set.

Definition $(3)$ implies Definition $(4)$
Let $H^-$ be the closure of $H$ by definition 3.

Let $x$ be a point in the boundary of $H$.

Suppose $x \notin H^-$.

Since $H^-$ is closed, by definition $S \setminus H^-$ is open in $T$.

Thus by definition $S \setminus H^-$ is a neighbourhood of $x$ which contains no points in $H$.

This contradicts the fact that $x$ is a point in the boundary of $H$.

Proof of (3) $\implies$ (4):

Proof of (4) $\iff$ (1):

First we show that $H^-$ is closed. Let $y\in T-H^-$, then it is not an isolated point of $H$, and it is also not a limit point of $H$. So there exists a neighbourhood $V_y$ of $H$ such that $V_{y}\cap (H-\{x\})=\emptyset$. So $H^-$ is closed and we have one inclusion. For the reverse inclusion, let $h$ be given. If it is an isolated point of $H$, we are done. If it is a limit point of $H$, then every neighbourhood $V_{h}$ of $h$ has non-empty intersection with $H-\{h\}$ and hence for every $K-\{h\}$. Since $K$ is closed, $h$ lies in every closed $K$ containing $H$.

$(2)$
This is proved in Set Closure is Smallest Closed Set.