Path-Connectedness is Equivalence Relation

Theorem
Let $T = \left({X, \vartheta}\right)$ be a topological space.

Let $a \sim b $ denote the relation:
 * $a \sim b \iff a$ is path-connected to $b$

where $a, b \in X$.

Then $\sim$ is an equivalence relation.

Proof
Checking in turn each of the criteria for equivalence:

Reflexivity
We define the constant mapping on $\R$:


 * $\forall x \in \R: f_a \left({x}\right) = a$

Thus:
 * $f_a \left({0}\right) = a$


 * $f_a \left({1}\right) = a$

From Constant Mapping is Continuous it follows that $a$ is path-connected to itself.

So $a \sim a$ and $\sim$ has been shown to be reflexive.

Symmetry
If $x \sim y$ then $x$ is is path-connected to $y$ by definition.

We form the mapping $g: \R \to \R$:
 * $g \left({x}\right) = 1 - x$

which is trivially continuous.

By Composition of Continuous Mappings is Continuous $g \circ f$ is continuous.

Putting it together we see that $g \circ f$ is the mapping which maps $0$ to $y$ and $1$ to $x$.

So $y \sim x$ 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.