Can $\mathbb{R}^2$ be homeomorphic to $\mathbb{R}\times\mathbb{R}$ with this topology?

Question

Consider $\mathbb{R}^2$ with the included point topology (open subsets are those containing $(0,0)$); and $\mathbb{R}$ with the same topology (open subsets are those containing $0$). Now consider the product topology $\mathbb{R}\times \mathbb{R}$ with each one having the topology explained before.Can $\mathbb{R}×\mathbb{R}$ and $\mathbb{R}^2$ be homeomorphic with this topology?

I can only think about trying to see if it is possible to have a continuous bijection between this two sets and see if it can be an open map, but I'm stuck on proving this, any help would be appreciated, thanks!

Answer 1

The open sets of R$^2$ are { U : (0,0) in U }.

Each open set of R×R is a union of sets of the form U×V where U,V are open with the special point 0 topology of R.
Thus (0,0) is in every open set of R×R.
Furthermore if (0,0) is in U then U is the union of sets for the form
{0}×{0,r}, {0,r}×{0}, r in R. Thus U is open within R×R.

Consequently, as R$^2$ and R×R have identical open sets, the identity map a homeomorphism.

Answer 2

A way to understand when a topology $\cal T$ in a product $X=X_1\times X_2$ is a product topology ${\cal T}_1\times{\cal T}_2$ is that we know the candidates to factor topologies: the projection $X\to X_1$ induces homeos $\{y=y_0\}\to X_1$, and similarly for the other projection. In our case, ${\cal T}|\{y=1\}$ is discrete and so ${\cal T}_1$ cannot be a point topology.