Path-Connectedness is Equivalence Relation

Theorem
Let $T = \left({S, \tau}\right)$ 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.

Symmetry
If $a \sim b$ then $a$ is is path-connected to $b$ by definition.

We form the mapping $g: \left[{0 \,.\,.\, 1}\right] \to \left[{0 \,.\,.\, 1}\right]$:


 * $g \left({x}\right) = 1 - x$

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.

Transitivity
Follows directly from Joining Paths makes Another Path.

$\sim$ has been shown to be reflexive, symmetric and transitive.

Hence by definition it is an equivalence relation.