Exercise 18.11

Question

Let $F: X 	imes Y 	o Z$.
  We say that $F$ is 	extbf{	extit{continuous in each variable separately}} if for each $y_0$ in $Y$, the map $h: X 	o Z$ defined by $h(x) = F(x 	imes y_0)$ is continuous, and for each $x_0 \in X$, the map $k: Y 	o Z$ defined by $k(y) = F(x_0 	imes y)$ is continuous.
  Show that if $F$ is continuous, then $F$ is continuous in each variable separately.

Dan Whitman · Accepted Answer

\begin{proof}
    To show that $F$ is continuous in $x$, consider any $y_0 \in Y$ and define $h: X 	o Z$ by $h(x) = F(x 	imes y_0)$.
    Now consider any $x \in X$ and any neighborhood $V$ of $h(x) = F(x 	imes y_0)$.
    Then $V$ is an open set containing $h(x) = F(x 	imes y_0)$ so that it is a neighborhood of $F(x 	imes y_0)$.
    Since $F$ is continuous, this means that there is neighborhood $U'$ of $x 	imes y_0$ in $X 	imes Y$ such that $F(U') \subset V$ by Theorem~18.1.
    It then follows that there is a basis element $U_X 	imes U_Y$ of $X 	imes Y$ containing $x 	imes y_0$ where $U_X 	imes U_Y \subset U'$.
    Since $X 	imes Y$ is a product topology, we have that $U_X$ is open in $X$ and $U_Y$ is open in $Y$.
    Then, since $x 	imes y_0 \in U_X 	imes U_Y$ we have that $x \in U_X$ and $y_0 \in U_Y$ so that $U_X$ is a neighborhood of $x$ in $X$.

So consider any $z \in h(U_X)$ so that $z = h(x')$ for some $x' \in U_X$.
    Then $x' 	imes y_0 \in U_X 	imes U_Y$ so that also $x' 	imes y_0$ in $U'$ since $U_X 	imes U_Y \subset U'$.
    It then also follows that $z = h(x') = F(x' 	imes y_0) \in F(U')$ so that $z \in V$ since $F(U') \subset V$.
    This shows that $h(U_X) \subset V$ since $z$ was arbitrary.
    It then follows that $h$ is continuous by Theorem~18.1.
    
    The proof that $F$ is continuous in $y$ is directly analogous.
  \end{proof}

Exercise 18.11

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Exercise 18.11

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