Prove that $ |X| \leq |Y| $ if $d(x) \geq d(y) \forall x,y \in E $ in a bipartite graph

Question

Prove that $ |X| \leq |Y| $ if $d(x) \geq d(y) \forall x,y \in E $ in a bipartite graph with equality valid if and only if $d(x) = d(y) \forall x,y \in E $ where d(v) stands for degree of vertex v

Why does it say $x,y \in E $ ? Aren't they vertices?

My approach:-

$$ |X| \leq d(y) \forall y$$ $$ |Y| \leq d(x) \forall x$$ And we have $$d(x) \geq d(y) \forall x,y $$ any help?

Answer 1

Since the question as stated doesn't make sense, I'm guessing that the real question is the following:

Consider a bipartite graph with partite sets $X,Y$ and edge set $E,$ and with no isolated vertices. Prove that, if $\operatorname{d}(x)\ge\operatorname{d}(y)$ whenever $x\in X,\ y\in Y,\ xy\in E$$,$ then $|X|\le|Y|,$ with equality only if $\operatorname{d}(x)=\operatorname{d}(y)$ for each edge $xy\in E.$

Proof: $$|X|=\sum_{xy\in E}\frac1{\operatorname{d}(x)}\le\sum_{xy\in E}\frac1{\operatorname{d}(y)}=|Y|$$ Explanation of the equalities: Recall that there are no isolated vertices. For each vertex $x\in X$ let $E_x$ be the set of all edges incident with $x.$ Then $$\sum_{xy\in E}\frac1{\operatorname d(x)}=\sum_{x\in X}\sum_{xy\in E_x}\frac1{\operatorname d(x)}=\sum_{x\in X}\left(\frac1{\operatorname d(x)}\cdot\sum_{xy\in E_x}1\right)=\sum_{x\in X}\left(\frac1{\operatorname d(x)}\cdot|E_x|\right)=\sum_{x\in X}1=|X|.$$