Basis for a topology that we will call the product topology I recently saw this exercise:
For any pair of topological spaces $(X,\tau_X)$ and $(Y,\tau_Y)$ consider the set $B=\{U\times V \mid  U\in \tau_X, V\in \tau_Y\}$. Prove that $B$ is a basis for a topology of $X\times Y$. We will call this topology the product topology.
Would it me correct if I proved that the open sets generated by this set is a topology?
Or do I have to prove that the union is $X\times Y$ and for every intersecting elements there is a third one that is contained on the intersection?
Thanks
 A: By definition, $B$ is a basis for a topology on $X\times Y$ if


*

*Each $(x,y) \in X\times Y$ is contained in some basis element in $B$;

*If $(x,y)$ belongs to two basis elements in $B$ then there exists a third basis element containing $(x,y)$ that is contained in the intersection of the two.


Then you may generate a topology on $X \times Y$ by taking unions of basis elements.
To wit that this really is a basis for a topology:


*

*The basis element $X \times Y$ clearly contains any point $(x,y)$;

*Suppose we have two basis elements $U_1 \times V_1$ and $U_2 \times V_2$ containing $(x,y)$. Then the intersection $(U_1 \times V_1) \cap (U_2 \times V_2) = (U_1 \cap U_2) \times (V_1 \cap V_2)$ is itself a basis element containing $(x,y)$.



Note that if you have an infinite number of factors instead of just two, this generates the so-called box topology. The product topology, although more natural in the end, has to be defined in a slightly more obscure way in that setting.
A: Let $X$ be a set and consider a subset $\mathscr{B}$ of $\mathscr{P}(X)$ (the power set of $X$) such that


*

*$\bigcup\mathscr{B}=X$;

*for each $U,V\in\mathscr{B}$, if $x\in U\cap V$, then there is $W\in\mathscr{B}$ such that $x\in W$ and $W\subseteq U\cap V$.


Then $\mathscr{B}$ is called a basis for a topology on $X$, with generated topology the set $\tau$ of arbitrary unions of elements of $\mathscr{B}$.
Verifying that $\tau$ is a topology on $X$ is straightforward.


*

*The empty set is the union of the empty subset of $\mathscr{B}$;

*$X\in \tau$ because of property 1 of a basis;

*Arbitrary unions of elements of $\tau$ are unions of unions of elements of $\mathscr{B}$, so they belong to $\tau$ as well

*The intersection of two elements of $\tau$ is an element of $\tau$.


Only the last part needs a detailed proof. Suppose $\mathscr{C}$ and $\mathscr{D}$ are subsets of $\mathscr{B}$ and set $U=\bigcup\mathscr{C}$ and $V=\bigcup\mathscr{D}$. We want to show that $U\cap V\in\tau$.
If $x\in U\cap V$, then $x\in U$ and $x\in V$; therefore there exist $B\in\mathscr{B}$ and $C\in\mathscr{C}$ such that $x\in B$ and $x\in C$. By definition of basis, there exists $W_x\in\mathscr{B}$ such that $x\in W_x$ and $W_x\subseteq B\cap C$. Evidently $W_x\subseteq U\cap V$ and therefore
$$
U\cap V=\bigcup\{W_x:x\in U\cap V\}
$$

So you don't need to show the properties above for the topology generated by the basis you're given, as they hold for every basis over every set.
You just need to show properties 1 and 2 of a basis.
Property 1 is trivial, because $X\in\tau_X$ and $Y\in\tau_Y$, so $X\times Y\in\mathscr{B}$ (which you denoted by $B$, but I'll stick to my notation above).
If $U_1,U_2\in\tau_X$, $V_1,V_2\in\tau_Y$ and $(x,y)\in(U_1\times V_1)\cap(U_2\times V_2)$, then clearly
$$
(x,y)\in(U_1\cap U_2)\times(V_1\cap V_2)\subseteq
(U_1\times V_1)\cap(U_2\times V_2)
$$
and, since $\tau_X$ and $\tau_Y$ are topologies,
$$
U_1\cap U_2\in\tau_X,\qquad V_1\cap V_2\in\tau_Y
$$
so
$$
(U_1\cap U_2)\times(V_1\cap V_2)\in\mathscr{B}
$$
