Ultraconnected Space is Path-Connected
Jump to navigation
Jump to search
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.
This article, or a section of it, needs explaining. In particular: The above is a nontrivial claim You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code. |
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