Exercise 18.10

Proof. Consider any $x\times y\in A\times C$ and any neighborhood $V$ of $(f\times g)(x\times y)$ in $B\times D$. Since $V$ is open in $B\times D$, there is a basis element $U_B\times U_D$ of the product topology that contains $(f\times g)(x\times y)$ where $U_B\times U_D\subset V$. Then $U_B$ and $U_D$ are open in $B$ and $D$, respectively.

Since $f$ is continuous, we then have that $U_A=f^{-1}(U_B)$ is open in $A$. Likewise $U_C=g^{-1}(U_D)$ is open in $C$ since $g$ is continuous. Then the set $U=U_A\times U_C$ is a basis element of the product topology $A\times C$ and therefore open. Now consider any $w\times z\in (f\times g)(U)$ so that there is an $x^{\prime }\times y^{\prime }\in U=U_A\times U_C$ where $w\times z=(f\times g)(x^{\prime }\times y^{\prime })=f(x^{\prime })\times g(y^{\prime })$. Hence $w=f(x^{\prime })$ and $x^{\prime }\in U_A=f^{-1}(U_B)$ so that $w=f(x^{\prime })\in U_B$. Similarly $z=g(y^{\prime })$ and $y^{\prime }\in U_C=g^{-1}(U_D)$ so that $z=g(y^{\prime })\in U_D$. Thus $w\times z\in U_B\times U_D$ so that also $w\times z\in V$ since $U_B\times U_D\subset V$. This shows that $(f\times g)(U)\subset V$ since $w\times z$ was arbitrary.

This suffices to show that $f\times g$ is continuous by Theorem 18.1 as desired. □