prove the continuity of a function

Question

Given two topological space $X$ and $Y$, $f:X\times Y\rightarrow \mathbb{R}$ is continuous. Then, how do we prove $f_x:Y\rightarrow\mathbb{R}$ defined as $f_x(y)\equiv f(x,y)$ is cotinuous for $\forall x\in X$?

For any open set $U\subset\mathbb{R}$, we need to show $f^{-1}_x(U)$ is open in $Y$. $f_x^{-1}(U)= \pi_Y(f^{-1}(U)\cap (\{x\}\times Y) )$, where $\pi_Y$ is the projection map, which is open. But $(\{x\}\times Y)$ is not open.

Ah, $f^{-1}(U)$ is open in $X\times Y$, thus, it is the union of basic open sets, i.e., $f^{-1}(U)=\cup_i A_i\times B_i$. Then, $\pi_Y(f^{-1}(U)\cap (\{x\}\times Y))=\cup_j B_j$ which is open in $Y$.

I would start by writing down the definition of what is assumed and what needs to be shown. — The Chaz 2.0, Oct 16 '20 at 18:01
You just need to show that $Y\to X\times Y:y\mapsto(x,y)$ is continuous — Kenta S, Oct 16 '20 at 18:05

score 0 · Accepted Answer · answered Oct 16 '20 at 21:32

Fix $x \in X$ and consider $i_x : Y \to X \times Y$ given by $i_x(y) = (x,y)$. Then, for an open set $U$ in $X \times Y$ we have $i_x^{-1}(U) = \pi_Y(U)$, which is open since the projections are open maps. It follows that $i_x$ is continuous, and hence, $f \circ i_x = f_x$ is also continuous.

prove the continuity of a function

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