Let $G=(V,E)$ be a finite undirected graph so vertex $x\in V$ has odd rank. I'm trying to prove that there must be vertex $y\in V$ which has odd rank and there is a path between $x,y$. I feel like it a basic theorem but I didn't find any proof for it and I can't seem to think of one.
How to approach this problem?
On
As explained by ajotatxe :
Given a connected component of a graph, the sum of the ranks of its vertices is precisely two times the number of edges, that is, an even number.
Let $H$ be the connected component including $x$, of odd degree : $d(x)=2k+1$. Let $m$ be the number of edges in $H$ : $$ \sum_{i \in V(H)} d(i) = 2m$$ $$ \sum_{i \in V(H)\backslash\{x\}} d(i) = 2m - d(x) = 2k'+1$$ Therefore there must exist $y \in V(H)\backslash\{x\}$ such that $d(y)$ is odd, and $y \in V(H)$ implies that there exist a path from $x$ to $y$.
Given a connected component of a graph, the sum of the ranks of its vertices is precisely two times the number of edges, that is, an even number.