# 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.

$\Box$

### 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.

$\Box$

### 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

- 1970: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*... (previous) ... (next): $\text{I}: \ \S 4$