Equivalence of Definitions of Closure of Topological Subspace

Theorem
Let $$T$$ be a topological space.

Let $$H \subseteq T$$.

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


 * 1) $$\operatorname{cl} \left({H}\right)$$ is the union of $$H$$ and its limit points;
 * 2) $$\operatorname{cl} \left({H}\right) = \bigcap_{H \subseteq K \subseteq T: K \text{ closed}} K$$;
 * 3) $$\operatorname{cl} \left({H}\right)$$ is the smallest closed set that contains $$H$$;
 * 4) $$\operatorname{cl} \left({H}\right)$$ is the union of $$H$$ and its boundary;
 * 5) $$\operatorname{cl} \left({H}\right)$$ is the union of all isolated points of $$A$$ and all limit points of $$H$$.

1 implies 2
Let $$V = \bigcap_{H \subseteq K \subseteq T: K \text{ closed}} K$$.

That is, $$V$$ is the intersection of all closed sets in $$T$$ that contain $$H$$.

Let $$K$$ be closed, and let $$H \subseteq K$$.

From statement 1: "$$\operatorname{cl} \left({H}\right)$$ is the union of $$H$$ and its limit points", these proofs have been demonstrated:
 * Closure of Subset is Subset of Closure;
 * Closed Set Equals its Closure;
 * Closure is Closed.

From Closure of Subset is Subset of Closure, we have $$\operatorname{cl}\left({H}\right) \subseteq \operatorname{cl}\left({K}\right)$$.

From Closed Set Equals its Closure, we have $$\operatorname{cl}\left({K}\right) = K$$.

So $$\operatorname{cl}\left({H}\right) \subseteq V$$.

Conversely, from Closure is Closed, $$\operatorname{cl}\left({H}\right)$$ is closed.

Hence $$V \subseteq \operatorname{cl}\left({H}\right)$$.

So $$V = \operatorname{cl}\left({H}\right)$$ and hence statement 1 implies statement 2.

2 implies 3
Let $$V = \operatorname{cl}\left({H}\right)$$ defined as in statement 2.

If $$K$$ is closed in $$T$$ and $$H \subseteq K$$, then $$V \subseteq K$$ from Intersection Subset.

So $$V = \operatorname{cl}\left({H}\right)$$ is a subset of any closed set in $$T$$ which contains $$H$$, and so is the smallest closed set that contains $$H$$.

3 implies 2
Let $$K$$ be closed set in $$T$$ such that $$H \subseteq K$$

If $$V$$ is the smallest closed set that contains $$H$$, then $$V \subseteq K$$.

It follows from Intersection Largest that $$V$$ is the intersection of all closed sets in $$T$$ that contain $$H$$.