Product topology. Clarification needed

Question

Please consider the statement and it's proof below:

The author further says that : Members of the product topology can all be expressed as union of products, but most members of the product topology are not products.

For example - the open disc $\{x \in \mathbb R^2~|~x_1^2+x_2^2 <1 \}$ is a member of the product topology on $\mathbb R^2$ but it is not a product of open intervals of $\mathbb R$. Infact, to express it as a union of products of open intervals, we need to use an infinite collection of such products in the following way: For each $x \in U$, where $U$ is an open set in the disc, set $r_x=1- \sqrt{x_1^2+x_2^2}$. Then $U$ can be expressed as the union $\bigcup \{(x_1-r_x~,~ x_1+r_x) \times (x_2-r_x~,~ x_2+r_x)~|~x \in U\}$

Is there a contradiction between the second para of theorem $4.5.1$ and the example cited above?

According to the second para, $W$ any open subset can be expressed as the union of products of open balls ( intervals here in case of $\mathbb R$ ). Then, the author refutes that this is not possible in the example.

I am confused, any help will be greatly appreciated. Thanks!

Answer 1

There is no contradiction. In that example, the author explains that:

• the open disk is an open set with respect to the product topology;
• it is not a product of two open intervals;
• it is an union of products of open intervals.

And all of this is consistent with the theorem that the author proved.