Prove that $B$ is basis of Topology on $X\times Y$

Question

Please check my proof; this proof is my first proof about topology.

I was asked to prove the following:

Let $B_{1}$ be basis of topology $T_{1}$, $B_{2}$ be basis of topology $T_{2}.$ The Set $X\times Y$ consist of all order pair $(x,y)$ with $x\in X$ and $y\in Y$. Let $B$ be the collection of all subsets of $X\times Y$ consisting of $B_{1}\times B_{2}$ where $B_{1} \in B_{1}$ and $B_{2}\in B_{2}$. Prove that $B$ is basis of a topology on $X\times Y$

1 Consider it satisfy first condition since $(x,y)\in B$
since all $x\in B_{1}$ and all $y\in B_{2}$ then $B$ is all collection of union of order pair that have component from $B_{1}$ and $B_{2}$

therefore it's satisfy first condition

2.condition 2

Suppose $C_{1}\subseteq B$ $C_{2}\subseteq B$ and let $(x_{1},y_{1})\in C_{1}$ and $(x_{1},y_{1})\in C_{2}$

there exist some $C_{3}$ that contain $(x_{1},y_{1})$

therefore it's satisfy second condition

Answer 1

Choose $(x,y)\in X×Y$ then there exists $U\in \mathscr B_1$ and there exists $V\in \mathscr B_2$ such that $x\in U\ and\ y\in V$. Then $(x,y)\in U×V \in \mathscr B$.

Choose two elements $U_1×V_1, U_2×V_2 \in \mathscr B$ both containing the element $(x_1,y_1)\in X×Y$ then $x_1\in U_1\cap U_2$ and $ y_1\in V_1\cap V_2$ , therefore there exists an element of $\mathscr B_1$ say, $ W_1(\subseteq U_1\cap U_2)$ containing $x_1$ and there exists an element of $\mathscr B_2$ say, $W_2(\subseteq V_1\cap V_2)$ containing $y_1$. Therefore $(x_1,y_1)\in W_1×W_2 \subseteq (U_1\cap U_2)×(V_1\cap V_2)$. Also we have $ W_1×W_2\in \mathscr B$.Note that $(U_1\cap U_2)×(V_1\cap V_2)=(U_1×V_1)\cap (U_2×V_2)$.

Answer 2

You check that $\mathcal{B}_1 \times \mathcal{B}_2$ is a base for some topology on $X \times Y$, but you have to check it is a base for the product topology, so for every $O$ open in the product topology and every $(x,y) \in O$ you have to find a set of the form $B_1 \times B_2, B_1 \in \mathcal{B}_1, B_2 \in \mathcal{B}_2$ such that $$(x,y) \in B_1 \times B_2 \subseteq O$$

This you can do because you first note that there must be an "open box" $O_1 \times O_2$ in $X \times Y$ with $O_1$ open in $X$ and $O_2$ open in $Y$, such that

$$(x,y) \in O_1 \times O_2 \subseteq O$$

by the definition of the product topology. Now we apply the fact that $\mathcal{B}_1$ is a base for $X$ to find $B_1 \in \mathcal{B}_1$ with $x \in B_1 \subseteq O_1$. Also, as $\mathcal{B}_2$ is a base for $Y$ we have $B_2 \in \mathcal{B}_2$ such that $y \in B_2 \subseteq O_2$.

Now $B_1 \times B_2$ is as required.