Set between Connected Set and Closure is Connected/Proof 2

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 $T_K = \struct {K, \tau_K}$ 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:

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

By hypothesis:

$K \subseteq H^-$

and so by Intersection with Subset is Subset‎:

$\map {\cl_K} H = K$

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

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $6.2$: Connectedness: Proposition $6.2.18$