0

Consider the family of graphs $A=\{\text{paths}\}$, $B=\{\text{cycles}\}$ and $D=\{\text{bipartite graphs}\}$. For each pair of these families, determine all isomorphism classes of graphs that belong to both families.

Since, a graph containing an odd cycle cannot form a bipartite graph, $B\cap D=$ even cycles $=\{C_{2k}:k\ge 1\}$. However, the solution given reads $B\cap D=\{C_{2k}:k\ge 2\}$. Why is that?

Also, for the part $A\cap D$, any path can be sectioned into two partite sets to form a partite graph but not all bipartite graphs form paths. Hence, $A\cap D=A$. But, the reason mentioned is non-bipartite graphs have odd cycles. What I know is that a graph with an odd cycle is non-bipartite. Of course, this does not imply that non-bipartite graphs have odd cycles. If this is true, could I get a rough proof?

zaira
  • 2,386
  • 2
  • 14
  • 37
  • 1
    They have $k\geq 2$ because there is no simple graph $C_2$ (this would be two vertices with two edges between them, which we normally disallow unless stated otherwise). So the smallest cycle is $C_3$. Your second question has been answered here:

    https://math.stackexchange.com/questions/311665/proof-a-graph-is-bipartite-if-and-only-if-it-contains-no-odd-cycles

     – Brandon du Preez May 25 '21 at 09:35
  • 1
    @BrandonduPreez completely slipped my mind that I was dealing with simple loopless graphs. Of course with two vertices, it is impossible to have more than 1 edge in a simple graph, unless at least 1 vertex is looped. The other doubt is also cleared, thanks! – zaira May 26 '21 at 06:59

0 Answers0