Equivalence of Definitions of Path Component/Equivalence Class equals Union of Path-Connected Sets
Jump to navigation
Jump to search
Theorem
Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in T$.
Let $\CC_x = \left\{ {A \subseteq S : x \in A \land A }\right.$ is path-connected in $\left. {T}\right\}$.
Let $C = \bigcup \CC_x$
Let $\sim$ be the equivalence relation defined by:
- $y \sim z$ if and only if $y$ and $z$ are path-connected in $T$.
Let $C'$ be the equivalence class of $\sim$ containing $x$.
Then $C = C'$.
Proof
\(\ds y \in C'\) | \(\leadstoandfrom\) | \(\ds x \text{ is path-connected to } y \text{ in } T\) | Definition of $\sim$ | |||||||||||
\(\ds \) | \(\leadstoandfrom\) | \(\ds \exists B \text{ a connected set of } T, x \in B, y \in B\) | Points are Path-Connected iff Contained in Path-Connected Set | |||||||||||
\(\ds \) | \(\leadstoandfrom\) | \(\ds \exists B \in \CC_x : y \in B\) | Equivalent definition | |||||||||||
\(\ds \) | \(\leadstoandfrom\) | \(\ds y \in \bigcup \CC_x\) | Definition of Union of Set of Sets | |||||||||||
\(\ds \) | \(\leadstoandfrom\) | \(\ds y \in C\) | Definition of $C$ |
The result follows.
$\blacksquare$