Arc-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 arc-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

The definition of arc-connected specifically makes provision that:

$\forall a \in S: a \sim a$

specifically to ensure that $\sim$ be reflexive.

$\Box$

Symmetry

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

In the trivial case of $x = y$, $y \sim x$ by reflexivity.

Suppose that $x \ne y$.

We form the mapping $g: \R \to \R$:

$\map g x = 1 - x$

which is trivially continuous and injective.

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

By Composite of Injections is Injection, $g \circ f$ is also injective.

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

So $y \sim x$ and $\sim$ has been shown to be symmetric.

$\Box$

Transitivity

Follows directly from Joining Arcs makes Another Arc.

$\Box$

$\sim$ 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