Difference between directed and undirected graph edge sets 
Hi guys, what is the difference exactly between the both edge sets V*V and [V]^2?
  


 A: $[V]^2$ is the family of unordered $2$-element subsets of $V$; $V\times V$ is set of ordered pairs of elements of $V$. 

Example: If $V=\{0,1,2\}$, then $$[V]^2=\big\{\{0,1\},\{0,2\},\{1,2\}\big\}\;,$$ corresponding to the three possible edges between vertices in $V$. But $$V\times V=\{\langle 0,0\rangle,\langle 0,1\rangle,\langle 0,2\rangle,\langle 1,0\rangle,\langle 1,1\rangle,\langle 1,2\rangle,\langle 2,0\rangle,\langle 2,1\rangle,\langle 2,2\rangle\}\;,$$ corresponding to the $9$ possible directed edges from a vertex of $V$ to a vertex of $V$. Note that in this case we allow loops, i.e., directed edges from a vertex to itself.

If $u$ and $v$ are distinct vertices, $\{u,v\}$ is an element of $[V]^2$ and potentially an edge of the undirected graph $G$. The same vertices can be used to form two different ordered pairs, $\langle u,v\rangle$ and $\langle v,u\rangle$; each of these is potentially an edge of the directed graph $G$, and they are different edges. You can think of $\langle u,v\rangle$ as an edge directed from the vertex $u$ to the vertex $v$, while $\langle v,u\rangle$ is an edge directed from the vertex $v$ to the vertex $u$; when direction matters (i.e., in a directed graph), these are different edges. In an undirected graph, on the other hand, we don’t care about direction, so we just specify the $2$-element set of endpoints of each edge.
A: In an undirected graph the edge $(a,b)$ is an arc or line joining vertices $a$ and $b$ without any direction. Moreover, $(a,b)$ and $(b,a)$ has same representation in graph.
In a directed graph the edge $(a,b)$ is an arc or line pointing from vertex $a$ to vertex $b$. Here $(b,a)$ is the reverse edge pointing from $b$ to $a$.
