Are the graphs with no vertex and 1 vertex bipartite? I am not sure how the definition of bipartite graph fits for these graphs. If they are bipartite where is the bipartition?
 A: A graph with no edges and 1 or n vertices is bipartite.
Mistake: It is very common mistake as people think that graph must be connected to be bipartite.
Correction: No it is not the case, as graph with no edges will be trivially bipartite.
Mistake: If graph has no circuits then it cannot be bipartite as all circuits must be of even length to make graph bipartite.
Correction: If graph have no circuits (means graph have 0 circuit so it also even number), so again by definition of bipartite graph it will be bipartite.
A: Wikipedia says:

A bipartite graph (or bigraph) is a graph whose vertices can be divided into two disjoint sets U and V such that every edge connects a vertex in U to one in V; that is, U and V are independent sets. Equivalently, a bipartite graph is a graph that does not contain any odd-length cycles.

By that definition (which matches the one I'd use, although I'm hardly an authority on such things), any graph with no edges is trivially bipartite.
And, yes, the bipartition of the empty graph consist of two empty sets — the empty set being the only set which is disjoint from itself, since its intersection with itself is empty.
Ps. It is a little known mathematical fact that all elements of the empty set are even, infinite, continuous, true and purple with yellow spots. :-)  (They are, of course, also many things besides those.)
A: Can you partition the set of vertices into two subsets (no overlap) such that there are no edges between vertices within either part? Note that a subset can be empty.
