Let $(X_1,\tau_1)$ and $(X_2,\tau_2)$ be separable spaces, and let $\tau$ denote the product topology on $X=X_1 \times X_2$. Then $(X,\tau)$ is a separable space.
Proof. $(X_1,\tau_1)$ and $(X_2,\tau_2)$ be separable spaces. So, there exists countable dense subsets $D_1$ of $(X_1,\tau_1)$ and $D_2$ of $(X_2,\tau_2)$ such that $\overline {D_1}=X_1$ and $\overline {D_2}=X_2$. Let $(x_1,y_1)\in X_1 \times X_2 \implies x_1 \in X_1 $ and $x_2 \in X_2$. Let $B\in \tau$ such that $(x_1,x_2)\in B$. So, there exists Let $U_1 \in \tau_1$ such that $x_1 \in U_1$. $U_1\cap D_1 \neq \phi$. Let $y_1 \in U_1\cap D_1 $. similarly, Let $U_2 \in \tau_2$ such that $x_2 \in U_2$. $U_2\cap D_2 \neq \phi$. Let $y_2 \in U_2\cap D_2 $. So, $(x_1,x_2) \in \overline {D_1 \times D_2}$. We know that $D_1 \times D_2$ countable. Hence $(X,\tau)$ is separable. Am I correct?
Best Answer
I will clarify some lines.
Let $W$ be non-empty open subset in $X_1×X_2$, choose an element $(x_1,x_2)\in W$. Now by definition of basis of product topology we have an open set $U\subseteq X_1$ containing $x_1$ and an open set $V\subseteq X_2$ containing $x_2$ such that $U×V\subseteq W$.
Now $\overline {D_1}=X_1$ and $\overline {D_2}=X_2$ so that $U\cap D_1\not=\phi,V\cap D_2\not=\phi$ . So that $$B\cap (D_1×D_2)\supseteq (U×V) \cap (D_1×D_2)=(U\cap D_1)×(V\cap D_2)\not=\phi$$Therefore $D_1×D_2$ is dense in $X_1×X_2$.