Ultraconnected Space is Path-Connected

Theorem

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

Then $T$ is path-connected.

Proof

Let $T = \struct {S, \tau}$ be ultraconnected.

Let $a, b \in S$.

Let $p \in \set a^- \cap \set b^-$ where $\set a^-$ is the closure of $\set a$.

Such a $p$ can be chosen, as $T$ being ultraconnected guarantees that $\set a^- \cap \set b^- \ne \O$.

Consider the mapping $f: \closedint 0 1 \to X$ such that:

$\map f x = \begin{cases}

a & : x \in \hointr 0 {\dfrac 1 2} \\ p & : x = \dfrac 1 2 \\ b & : x \in \hointl {\dfrac 1 2} 1 \\ \end{cases}$

Then $f$ is continuous.

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The result follows from the definition of path-connectedness.

$\blacksquare$

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