For an open subset $U\subseteq X\times Y$, is the section $U_x$ open in $Y$?

Question

Let $X,Y$ be topological spaces and let $U$ an open subset of $X\times Y$. For $x\in X$, let $$U_x=\{y\in Y: (x,y)\in U\}$$

Is it $U_x$ an open subset of $Y$?

Thanks.

Answer 1

You have $$U_x=\{y\in Y:(x,y)∈U\}=\pi_Y(U\cap\{x\}\times Y)$$ where $\pi_Y$ is the projection onto $Y$. Or, if $\barπ$ denotes the restriction of $π_Y$ to $\{x\}×Y$, then $U_x=\barπ(U∩\{x\}×Y)$. Can you show that $U∩\{x\}×Y$ is open in $\{x\}×Y$ and that $\barπ$ is a homeomorphism between its domain and $Y$?

Answer 2

Clearly, $U_x=\pi_Y^{-1}(U\cap(\{x\}\times Y))$.

Since $U$ is open in $X\times Y$ there is a basic subset $G$ such that $p\in G\subset U$.

For this basic $G$ there exist two open sets $A\subset X$ and $B\in Y$ respectively, such that $$G=\pi_X^{-1}A\cap\pi_Y^{-1}B.$$

We will prove that $p\in B\subset U_x$.

For if $q\in B$ then $(x,q)\in\pi_Y^{-1}B$. But also $(x,q)\in\pi_X^{-1}A$, since $\pi_X(x,q)=x$.

So $(x,q)\in G$ and hence $(x,q)\in U$.

This proves $p\in B\subset U_x$ and hence $U_x$ is open.