Slices of a Continuous Function are Continuous

Last updated Nov 1, 2022

Let $X,Y,Z$ be Topological Spaces and suppose $F: X \times Y \to Z$ is a Continuous Function. Then $F_{x}$, is a Continuous Function.

Suppose $V \subset Z$ is Open. Then $$\begin{align*} &y \in F_{x}^{-1}(V)\\ \Leftrightarrow& F(x, y) \in V\\ \Leftrightarrow&(x,y) \in F^{-1}(V)\\ \Leftrightarrow&(x,y) \in F^{-1}(V) \cap ({x} \times Y).\\ \Leftrightarrow&y \in \pi_{2}(F^{-1}(V) \cap ({x} \times Y)), \end{align*}$$ so $F^{-1}{x}(V) = \pi{2}(F^{-1}(V) \cap ({x} \times Y))$. Since $F^{-1}(V)$ is Open in $X \times Y$ and Product of Finitely Many Proper Open Sets form a Basis for the Product Topology, we know $$F^{-1}(V) = \bigcup\limits_{i \in I} U_{X,i} \times U_{Y,i}.$$ for Open sets $U_{X,i} \subset X$ and $U_{Y, i} \subset Y$ for $i \in I$. Then $$\pi_{2}(F^{-1}(V) \cap ({x} \times Y)) = \pi_{2} ( \bigcup\limits_{i \in I} (U_{X,i} \times U_{Y,i} \cap {x} \times Y) = \pi_{2} ( \bigcup\limits_{i \in I} {x} \times U_{Y,i}) = \bigcup\limits_{i \in I} U_{Y,i},$$ which is Open in $Y$. $\blacksquare$

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