Let $$G=(V,E)$$ be a complete bipartite graph of an order at least two with partite sets X and Y. Then either $$|X|=1$$ or $$|Y|=1$$.

Proof

Assume G is a tree and $$|Y|>1$$. We prove the statement by contradiction.

Suppose $$|X|>1$$. Pick any pair of distinct vertices $$x_1$$ and $$x_2$$ in X, and follow any edge from $$x_1$$ to some vertex $$y_1$$ in Y. There must be such an edge $$\{y_1,x_2\}\in E(G)$$ since G is a complete bipartite graph. Furthermore, there must be another vertex $$y_2$$ in Y since $$|Y|>1$$. Because G is a complete bipartite graph, we have $$\{x2,y2\}\in E(G)$$ and $$\{y2,x1\}\in E(G)$$. Now consider a path $$x_1{y}_1{x}_2{y}_2{x}_1$$, which is a cycle. But G is a tree so G is acyclic and this is a contradiction. Thus our supposition must be false and $$|X|=1$$. A similar argument shows that $$|Y|=1$$ if $$|X|>1$$.

Q.E.D.