let $\tau$ denote the product topology on $X=X_1 \times X_2$. Then $(X,\tau)$ is a first countable.

Question

Let $(X_1,\tau_1)$ and $(X_2,\tau_2)$ be first countable, and let $\tau$ denote the product topology on $X=X_1 \times X_2$. Then $(X,\tau)$ is a first countable.

Proof. Let $(x,y)\in X_1 \times X_2$ be any arbitrary. We have there exists a countable local basis($\mathscr {B_x}$(say)) at $x$ and countable local basis($\mathscr {B_y}$(say)) at $y$. $\mathscr B_{(x,y)}=\{B_1 \times B_2:B_1 \in \mathscr {B_x} $ and $B_2 \in \mathscr {B_y}\}$ is the desired countable local basis at $(x,y)$. Am I correct? Let $(x,y)\in B \in\tau$. so there exists a $U\times V$, $U\in \tau_1$ and $V\in \tau_2$. so there exists $B_x \in \mathscr {B_x}$ and $B_y \in \mathscr {B_y}$. So, I got an element of $ \mathscr B_{(x,y)}$ containing $(x,y)$ which sits inside $B$. Hence, $\mathscr B_{(x,y)}$ is a countable local basis at $(x,y)$. Am I proved correctly? Please help me.

Answer 1

I clearify some lines of your proof.

Let $B$ be a open nbd of $(x,y)\in X×Y$ , then by definition of basis of product topology we have a open nbd $U_x$ of $x$ and open nbd $V_y$ of $y$ such that $U_x×V_y\subseteq B$. Now using definition of local base we have $B_1\in \mathscr B_x$ and $B_2\in \mathscr B_y$ such that $B_1\subseteq U_x$ and $B_2\subseteq V_y$. Therefore $B_1×B_2\subseteq U×V\subseteq B$. Hence $\mathscr B_{(x,y)}$ is a local base at $(x,y)\in X×Y$.