Set between Connected Set and Closure is Connected/Proof 1

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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.


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:

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


$k \in H^-$

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.