Graph Connectedness is Equivalence Relation

Theorem
Let $$G = \left({V, E}\right)$$ be a graph.

Let $$\to$$ denote the relation "is connected to" on the set $$V$$.

Then $$\to$$ is an equivalence relation.

Proof
Let $$u, v, w$$ be arbitrary vertices of a graph $$G$$.

Checking in turn each of the criteria for equivalence:

Reflexive
By definition, all vertices are connected to themselves.

Hence $$\to$$ is reflexive.

Symmetric
Suppose $$u \to v$$. Then by definition there exists a walk $$W$$ from $$u$$ to $$v$$.

Let $$W = \left({u, u_1, u_2, \ldots, u_{k-1}, u_k, v}\right)$$.

Any edge $$\left\{{u_i, u_j}\right\}$$ can be traversed in either direction on a walk.

So $$W' = \left({v, u_k, u_{k-1}, \ldots, u_2, u_1, u}\right)$$ is also a walk.

Hence $$v \to u$$, and so $$\to$$ is symmetric.

Transitive
Suppose $$u \to v$$ and $$v \to w$$.

Let:
 * $$W_1 = \left({u, u_1, u_2, \ldots, u_{k-1}, u_k, v}\right)$$ be a walk from $$u$$ to $$v$$;
 * $$W_2 = \left({v, v_1, v_2, \ldots, v_{k-1}, v_k, w}\right)$$ be a walk from $$v$$ to $$w$$.

Then $$W = \left({u, u_1, u_2, \ldots, u_{k-1}, u_k, v, v_1, v_2, \ldots, v_{k-1}, v_k, w}\right)$$ is a walk from $$u$$ to $$w$$.

Hence $$u \to w$$, and so $$\to$$ is transitive.

Thus "is connected to" is an equivalence relation.