Path-Connectedness is Equivalence Relation
Theorem
Let $T = \struct {S, \tau}$ be a topological space.
Let $a \sim b $ denote the relation:
- $a \sim b \iff a$ is path-connected to $b$
where $a, b \in S$.
Then $\sim$ is an equivalence relation.
Proof
Checking in turn each of the criteria for equivalence:
Reflexivity
From Point is Path-Connected to Itself, we have that $a \sim a$.
So $\sim$ is reflexive.
$\Box$
Symmetry
If $a \sim b$ then $a$ is is path-connected to $b$ by definition.
We form the mapping $g: \closedint 0 1 \to \closedint 0 1$:
- $\map g x = 1 - x$
which is trivially continuous.
By Composite of Continuous Mappings is Continuous $f \circ g$ is continuous.
Putting it together we see that $f \circ g$ maps $0$ to $b$ and $1$ to $a$.
So $b \sim a$ and $\sim$ has been shown to be symmetric.
$\Box$
Transitivity
Follows directly from Joining Paths makes Another Path.
$\Box$
$\sim$ has been shown to be reflexive, symmetric and transitive.
Hence by definition it is an equivalence relation.
$\blacksquare$
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