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
Aiming for a contradiction, suppose $H$ 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 $V = \ds \bigcup_{y \mathop \in H \\ y \mathop \ne x} V_y$.
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 by Proof by Contradiction it must be the case that $H$ is dense-in-itself.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $4$: Connectedness: Disconnectedness