I was reading "Graph Theory" by Murty and Bondy and came across the following theorem which confuses me a bit.
In the line which I've highlighted by red color it seems that they are using the identity $d_G(x)=\#\{y\in V: xy\in E\}$, where $d_G(x)$ is the degree of vertex $x$.
But I guess that this formula is false. Take a look at the below picture.
We see that $d(1)=4$ but $\{y\in V: 1y\in V\}=\{2,4\}$ and hence $\#\{y\in V: 1y\in V\}=2$. Am I missing something here?
The fact that $d_G(x)=\#\{y\in V: xy\in E\}$ is definitely false and you already provided an example.
So you know that $G=G[X,Y]$ is a bipartite graph with $X=\{x_k\}_{k=1}^n$ and $Y=\{y_k\}_{k=1}^{m}$. Let's coonsider adjacency matrix $B=(b_{ij})$ of size $n\times m$, where $b_{ij}$ denotes the number of edges joining $x_i$ and $y_j$. One can see that $d_G(x_i)=b_{i1}+\dots+b_{im}$ for each $1\leq i\leq n$. The last identity means that sum of elements of $i$th row is $d_G(x_i)$.
Let's consider the new matrix $\tilde{B}=(\tilde{b}_{ij})$ obtained from $B$ in the following way: the $i$-th row of matrix $B$ we divide to $d_G(x_i)$ (and you see that $d_G(x_i)\neq 0$ because $G$ without isolated points). Matrix $\tilde{B}$ has the following interesting property: the sum of elements in each row is $1$. Hence the sum of all entries of $\tilde{B}$ is $n$, i.e. $$\sum \limits_{i=1}^{n}\sum \limits_{j=1}^{m}\tilde{b}_{ij}=\sum \limits_{i=1}^{n}1=n=|X|.$$
But let's take a look at the above sum from a different angle: $$|X|=\sum \limits_{i=1}^{n}\sum \limits_{j=1}^{m}\tilde{b}_{ij}=\sum \limits_{j=1}^{m}\sum \limits_{i=1}^{n}\tilde{b}_{ij}=\sum \limits_{j=1}^{m}\sum \limits_{i=1}^{n}\dfrac{b_{ij}}{d_G(x_i)};$$
But $\dfrac{b_{ij}}{d_G(x_i)}\leq \dfrac{b_{ij}}{d_G(y_j)}$ (If $b_{ij}=0$, then we have equality. If $b_{ij}>0$, then $d_{G}(x_i)\geq d_G(y_j)$). Hence $$|X|\leq \sum \limits_{j=1}^{m}\sum \limits_{i=1}^{n}\dfrac{b_{ij}}{d_G(y_j)}.$$ However, the inner sum in the last summation is the sum of all entries of $j$-th row of matrix $\hat{B}$ obtained from $B$ in the following way: we divide $j$-th column of $B$ by $d_G(y_j)$. We know that the sum of entries in each column is $1$. Hence $$|X|\leq \sum \limits_{j=1}^{m}\sum \limits_{i=1}^{n}\dfrac{b_{ij}}{d_G(y_j)}= \sum \limits_{j=1}^{m}1=m=|Y|.$$