Connected Component is not necessarily Open

Theorem

Let $T = \struct {S, \tau}$ be a topological space.

Let $H \subseteq T$ be a connected component of $T$.


Then it is not necessarily the case that $H$ is open.


Proof

Let $T = \struct {\Q, \tau_d}$ be the rational number space formed by the rational numbers $\Q$ under the usual (Euclidean) topology $\tau_d$.

Let $x \in \Q$.

From Components of Rational Number Space are Singletons, $\set x$ is a connected component of $T$.



This needs considerable tedious hard slog to complete it.
In particular: Need to establish that $\set x$ is not an open set of $\Q$.
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Sources

  • 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): connected component
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