A basis for topology of $\mathbb R^2$

Question

I'm studying via John Lee's book on general toplogy and i've came across this question, but it doesn't makes any sense to me, unless the following basis is a basis for product topology:

But is there any other topology that could be generated?

Answer 1

That is not a basis for the product topology, or indeed any other topology you're familiar with on the plane. The sets in $\mathcal{B}$ are vertical line segments, while the usual basis for the product topology would be open rectangles - and the product topology is the usual topology on $\mathbb{R}^2$.

Now for those identity maps, we're comparing this to the usual topology. If a set is open relative to $\mathcal{B}$, is it open in the usual topology? If a set is open in the usual topology, is it open relative to $\mathcal{B}$?

Answer 2

This is essentially the product topology of $\mathbb{R}$ in the discrete topology in the first factor, and the usual topology in the second. So it's clearly strictly finer than the usual (=product) topology on $\mathbb{R}^2$, so that the identity is only continuous with the usual topology as co-domain.