Set between Connected Set and Closure is Connected

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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 $T_K = \left({K, \tau_K}\right)$ be the topological subspace of $T$ whose underlying set is $K$.

Let $\operatorname{cl}_K \left({H}\right)$ denote the closure of $H$ in $K$.

From Closure of Subset in Subspace:

$\operatorname{cl}_K \left({H}\right) = K \cap H^-$

By hypothesis:

$K \subseteq H^-$

and so by Intersection with Subset is Subset‎:

$\operatorname{cl}_K \left({H}\right) = K$


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

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:

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


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

From Continuity Defined by Closure:

$f \left({\operatorname{cl}_K \left({H}\right)}\right) \subseteq \operatorname{cl}_K \left({f \left({H}\right)}\right) = \left\{{0}\right\}^-$

where $\left\{{0}\right\}^-$ is the closure of $\left\{{0}\right\}$ in $D$.

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

Thus by Set is Closed iff Equals Topological Closure:

$\left\{{0}\right\}^- = \left\{{0}\right\}$

That is, $f \left({K}\right) = \left\{{0}\right\}$.

Thus $K$ is connected by definition.

$\blacksquare$


Sources