Set between Connected Set and Closure is Connected/Proof 1

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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.


Proof

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

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:

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


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

Aiming for a contradiction, suppose $\exists k \in K: \map f k = 1$.

By definition of discrete space, $\set 1$ is open in $D$.

Hence by definition of continuous mapping:

$f^{-1} \sqbrk {\set 1}$ is open in $K$.

Let $K$ be given the subspace topology.

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

$f^{-1} \sqbrk {\set 1} = K \cap U$

We have that:

$k \in f^{-1} \sqbrk {\set 1} \subseteq U$

and:

$k \in H^-$

By definition of topology:

$\exists x \in H \cap U$

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

$\map f x = 0$

But because $x \in H \cap U \subseteq K \cap U = f^{-1} \sqbrk {\set 1}$:

$\map f x = 1$

This contradicts the definition of mapping.

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

Thus $K$ is connected.

$\blacksquare$


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