Ah, let $X,Y$ be top. spaces with $Y$ compact and $N$ open with $\{x\}\times Y\subseteq N$. For each $y\in Y$, there is a simple neighborhood $(x,y)\in U_y\times V_y\subseteq N$. Then $\{V_y\}_y$ is an open cover of $Y$, hence admits a finite subcover $V_{y_1},\dotsc,V_{y_n}$.
Then, $U=U_{y_1}\cap\dotsc\cap U_{y_n}$ is open. If $(x,y)\in U\times Y$, there is an index $i$, such that $y\in V_{y_i}$ (since the $V_{y_i}$ cover $Y$) and $x\in U_{y_i}$; so $(x,y)\in U_{y_i}\times V_{y_i}\subseteq N$, hence $U\times Y\subseteq N$.
Prove or disprove:
If $\mathcal{R}$ is a relation then
$\lvert\mathcal{R}\rvert=\lvert\mathcal{R}^{-1}\rvert$.
I think it is true but I do not know how to prove it.
Facts:
$\mathcal{R}^{-1}=\{(y,x)\mid(x,y)\in\mathcal{R}\}$.
$\mathcal{R}\subseteq A\times B$.
$\lvert\mathcal{R}\rvert\...
