Proving that $\mathcal B$ is a basis for a product topology

Question

In my general topology textbook there is the following exercice:

Let $\mathcal B_1$ be a basis for a topology $\tau_1$ on a set $X$, and $\mathcal B_2$ be a basis for a topology $\tau_2$ on a set $Y$. Let $\mathcal B$ be a be a collection of subsets $X \times Y$ consisting of all the sets $B_1 \times B_2$, where $B_1 \in \mathcal B_1$ and $B_2 \in \mathcal B_2$. Prove that $\mathcal B$ is a basis for a topology on $X \times Y$.

My first thought was to use the following properties:

$\mathcal B$ is a basis for some topology on a set $X$ if an only if:

• $X = \bigcup_{B \in \mathcal B} B$
• for $B_1,B_2 \in \mathcal B$, then the $B_1 \cap B_2$ is the union of members of $\mathcal B$

I think that we can use this because we already know that $\mathcal B_1$ and $\mathcal B_2$ satisfy these properties, but I'm still confused with this exercise, and I don't want so solve it just by using some properties and not understanding the problem. Can someone give me some tips or some intuition behind this problem that might help me understand it? Anything helps really, I'm just having some trouble "visualizing" the problem

Answer 1

You may visualize your problem by considering the case $X=Y=\mathbb{R}$. We may infer our exercise from this basic case. Connecting the visualization and the real proof would be another point, however.

I do not know there is any further intuition behind this problem. Sometimes mathematics is 'just do and works,' although your exercise is what everyone desires: we can find a basis of a given product of spaces by taking a product of two bases.

---

Now let us see how to solve your problem. Your attempt is nice, and it works with some modifications. The following description of the second condition on $\mathcal{B}$ would be helpful:

For any $U, V\in\mathcal{B}$ and $x\in U\cap V$, there is $N\in\mathcal{B}$ such that $x\in N\subseteq U\cap V$.

Moreover, we have to check the collection of open sets generated by $\mathcal{B}$ is exactly same as the collection of open sets of $X\times Y$.

Answer 2

Hint: the first property should be easy. For the second, it is enough to show the following: if $B_1,B_2\in \mathcal B$ and $x\in B_1\cap B_2$, then there is some $B_3\in \mathcal B$ such that $x\in B_3\subseteq B_1\cap B_2$.