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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?

S.Dan
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    What do you mean with $d$? And are $X,Y$ finite sets? – Stefan Mesken Jun 15 '15 at 01:35
  • d(v) stands for degree of vertex v. In this case, X and Y are two sets of vertices; each edge of a bipartite graph has one end in X and the other end in Y - definition of a bipartite graph. – S.Dan Jun 15 '15 at 02:17
  • Ah, I see. I thought $d$ was some kind of distance function. – Stefan Mesken Jun 15 '15 at 02:18
  • What textbook is this? It might just be that the $E$ is just a typo - I'd see if the publisher/author has released any errata. –  Jun 15 '15 at 02:20
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    Perhaps "$x,y \in E$ is not supposed to mean $x \in E$ and $y \in E$" but rather "the ordered pair $(x,y)$ is in $E$". – Gregory J. Puleo Jun 15 '15 at 03:23

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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 $$\sum_{xy\in E}\frac1{\operatorname d(x)}=\sum_{x\in X}\sum_{xy\in E_x}\frac1{\operatorname d(x)}=\sum_{x\in X}\left(\frac1{\operatorname d(x)}\cdot\sum_{xy\in E_x}1\right)=\sum_{x\in X}\left(\frac1{\operatorname d(x)}\cdot|E_x|\right)=\sum_{x\in X}1=|X|.$$

bof
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