Non-Trivial Connected Set in T1 Space is Dense-in-itself

Theorem
Let $T = \struct {S, \tau}$ be a $T_1$ (Fréchet) topological space.

Let $H \subseteq S$ be connected in $T$.

If $H$ has more than one element, then $H$ is dense-in-itself.

Proof
$H$ is not dense-in-itself.

Then $\exists x \in H$ such that $x$ is isolated in $H$.

That is, $\exists U \in \tau: U \cap H = \set x$.

Since $H$ has more than one element, we can find $y \in H$ with $y \ne x$.

Since $T$ is $T_1$:


 * $\forall y \in H: y \ne x \implies \paren {\exists V_y \in \tau: y \in V_y, x \notin V_y}$

Define $V = \displaystyle \bigcup_{y \in H\\ y \ne x} V_y$.

Then $\forall z \in H$:


 * $z = x \implies z \in U$
 * $z \ne x \implies z \in V_z \subseteq V$

Showing that $H \subseteq U \cup V$.

This also shows that both $U \cap H$ and $V \cap H$ are non-empty.

Since $x \notin V_y$ for any $y \in H$,

we have $x \notin V$.

This implies $\set x \cap V = \O$

So $H \cap U \cap V = \O$.

Thus we have:


 * $H \subseteq U \cup V$
 * $H \cap U \cap V = \O$
 * $U \cap H \ne \O$
 * $V \cap H \ne \O$

Showing that $H$ is disconnected.

This is a contradiction.

Hence we must have $H$ is dense-in-itself.