Show that ~ is an equivalence relation on the graph properties. (Discrete math) We're learning about Isomorphism, relations on graphs and graphs in general. I read this question in the book and this was the proof:
Clearly, a connected graph has a single component. On the
other hand, any two vertices x, y in the same component of a graph
G can be connected by a walk. Any walk from x to y of the shortest
possible length must be a path.
I don't understand it 100% because i know that equivalence relations from Logic and set theory have to be reflexive, Symmetric and transitive but here it has another meaning. So can someone explain me both the proof and the relation on graphs?

 A: Here it has the same meaning. You are defining a relation on vertices of a (i assume undirected) graph.
Let $v \in V$. We have that $v$~$v$ since we have a path from a vertex to itself (the empty path). So i t is reflexive.
Let $v, u \in V$. If $v$~$u$ Then there is a path from $v$ to $u$. Inversely, if you walk that path from end to start, you get a path from $u$ to $v$, so $u$~$v$. So it is symmetric.
****Edit**: I wish to stress that this is only correct for undirected graphs.** In a directed graph it is possible to have a path from $u$ to $v$ without having a path from $v$ to $u$ and this fails.
Now let $v, u, w \in V$ such that there is a path from $v$ to $u$ and from $u$ to $w$. This trivially means that there is a path from $v$ to $w$. Just go to $u$ and from there go to $w$. So $v$~$w$ which means transitivity.
So this is a equivalence relation.
Let's try to find the classes.
What is $[v]$? It is the set of all $u$ in $V$ such that there is a path from $v$ to $u$. this is just another way of saying "the component in which $v$ is in."
