In many topology books, the product topology of $X\times Y$ is defined via a basis, that is a topology generated by the basis $$\mathcal{B}:=\{U\times V:U\ \text{open in}\ X, V\ \text{open in}\ Y\}.$$
However, what if we assume NO knowledge of basis, but define a collection of sets $\mathcal{T}$ consisting of unions of sets of the form $U\times V$, where $U$ open in $X$ and $V$ open in $Y$?
Then, to show $\mathcal{T}$ is indeed a topology, the only thing we can use are those three axioms of being a topology. Then, how can I show $\mathcal{T}$ is a topology?
Then it is by definition that $\bigcup_{i\in I}(U\times V)_{i}\in \mathcal{T}$, since $\mathcal{T}$ contains any unions of sets of this form.
But how to show $X\times Y$, $\varnothing$ are open? and how to show $\mathcal{T}$ is closed under finite intersection?
It seems that we need to express everything into the form of the union of sets of the the product of open sets..
The set $X\times Y$ is open because both $X$ and $Y$ are open.
The empty set is open because it is equal to $\emptyset\times\emptyset$.
Finally, if $U_1$ and $U_2$ are open subset of $X$ and $V_1$ and $V_2$ are open subset of $Y$, then$$(U_1\times V_1)\cap(U_2\times V_2)=(U_1\cap U_2)\times(V_1\cap V_2).$$Now, if $A,B\in\mathcal T$, then$$A=\bigcup_{\lambda\in\Lambda_1}U_\lambda\times V_\lambda\text{ and }B=\bigcup_{\lambda\in\Lambda_2}U_\lambda\times V_\lambda.$$And then you use the fact that$$A\cap B=\bigcup_{(\lambda_1,\lambda_2)\in\Lambda_1\times\Lambda_2}(U_{\lambda_1}\times V_{\lambda_1})\cap(U_{\lambda_2}\times V_{\lambda_2}).$$