# Set between Connected Set and Closure is Connected

## Theorem

Let $T$ be a topological space.

Let $H$ be a connected set of $T$.

Let $H \subseteq K \subseteq H^-$, where $H^-$ denotes the closure of $H$.

Then $K$ is connected.

### Corollary 1

Let $T$ be a topological space.

Let $H$ be a connected set of $T$.

Let $H^-$ denote the closure of $H$ in $T$.

Then $H^-$ is connected in $T$.

### Corollary 2

Let $T = \struct{S, \tau}$ be a topological space.

Then every connected component of $T$ is closed.

## Proof 1

Let $D$ be the discrete space $\left\{{0, 1}\right\}$.

Let $f: K \to D$ be an arbitrary continuous mapping.

From Continuity of Composite with Inclusion, the restriction $f \restriction_H$ is continuous.

We have that:

- $H$ is connected
- $f \restriction_H$ is continuous

Thus by definition of connected set:

- $f \left({H}\right) = \left\{{0}\right\}$ or $f \left({H}\right) = \left\{{1}\right\}$

Without loss of generality, let $f \left({H}\right) = \left\{{0}\right\}$.

Aiming for a contradiction, suppose $\exists k \in K: f \left({k}\right) = 1$.

By definition of discrete space, $\left\{{1}\right\}$ is open in $D$.

Hence by definition of continuous mapping:

- $f^{-1} \left({\left\{{1}\right\}}\right)$ is open in $K$.

Let $K$ be given the subspace topology.

Then for some $U$ open in $T$:

- $f^{-1} \left({\left\{{1}\right\}}\right) = K \cap U$

We have that:

- $k \in f^{-1} \left({\left\{{1}\right\}}\right) \subseteq U$

and:

- $k \in H^-$

By definition of topology:

- $\exists x \in H \cap U$

As $x \in H$, we have that:

- $f \left({x}\right) = 0$

But because $x \in H \cap U \subseteq K \cap U = f^{-1} \left({\left\{{1}\right\}}\right)$:

- $f \left({x}\right) = 1$

This contradicts the definition of mapping.

Thus by Proof by Contradiction, $f: K \to D$ can not be a surjection.

Thus $K$ is connected.

$\blacksquare$

## Proof 2

Let $T_K = \struct {K, \tau_K}$ be the topological subspace of $T$ whose underlying set is $K$.

Let $\map {\cl_K} H$ denote the closure of $H$ in $K$.

From Closure of Subset in Subspace:

- $\map {\cl_K} H = K \cap H^-$

- $K \subseteq H^-$

and so by Intersection with Subset is Subsetâ€Ž:

- $\map {\cl_K} H = K$

Let $D$ be the discrete space $\set {0, 1}$.

Let $f: K \to D$ be any continuous mapping.

From Continuity of Composite with Inclusion, the restriction $f \restriction_H$ is continuous.

SWe have that:

- $H$ is connected
- $f \restriction_H$ is continuous

Thus by definition of connected set:

- $f \sqbrk H = \set 0$ or $f \sqbrk H = \set 1$

Without loss of generality, let $f \sqbrk H = \set 0$.

From Continuity Defined by Closure:

- $f \sqbrk {\map {\cl_K} H} \subseteq \map {\cl_K} {f \sqbrk H} = \set 0^-$

where $\set 0^-$ denotes the closure of $\set 0$ in $D$.

As $D$ is the discrete space, it follows from Set in Discrete Topology is Clopen that $\set 0$ is closed in $D$.

Thus by Set is Closed iff Equals Topological Closure:

- $\set 0^- = \set 0$

That is, $f \sqbrk K = \set 0$.

Thus $K$ is connected by definition.

$\blacksquare$

## Sources

- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*(2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $4$: Connectedness