TITLE: Bipartite subgraph having minimum degree
QUESTION [5 upvotes]: Prove that every graph $G$ with minimum degree $2m$ will have a bipartite subgraph having minimum degree $m$.
I have tried this proof by first taking one vertex $v$ and consider it in set $D_0$. Then i am taking m of its neighbours (out of $2m$) to form a set $D_1$. My main intension is to partition all vertex in G into $D_i$ s and create two partitions with $i$s as even in one partiotion and $i$s as odd in another partition.
But i am having a difficulty in proving that no edge exists between two $D_i$ in different levels

REPLY [3 votes]: Hint:

Let $G_0$ be any maximal bipartite subgraph of $G$.
Let $v_i \in V$ be some vertex of the smallest degree in $G_i$.
If $\mathrm{deg}_{G_i}(v_i) \geq m$, then you are done (and the sequence stops), otherwise set $G_{i+1}$ to be the graph with $v_i$ moved to the other side (and appropriate edges added or removed so that $G_{i+1}$ is also a maximal bipartite subgraph of $G$).
Observe, that with each turn, the number of edges of $\{G_i\}_i$ strictly increases, but the only way the sequence could stop is when $\mathrm{deg}_{G_\text{end}}(u) \geq m$ for all $u \in V$.

I hope this helps $\ddot\smile$