Q:

Crosses topology

At the plane $X=\mathbb{R^2}$ we consider the collection $\mathcal{T}$ of subsets $U\subset X$ such that for all $(a,b)\in U$ there exists $\epsilon>0$ with
$$((a-\epsilon,a+\epsilon)\times\{b\})\cup(\{a\}\times(b-\epsilon,b+\epsilon))\subset U$$
It's easy to see that $\mathcal{T}$ is a topology on $\mathbb{R}^2$. Also, every open set in the usual topology of $\mathbb{R}^2$ is also an open set in this topology ( because in every open ball we can  contain a "cross" as the ones defined above) and no points of $X$ are open, so $\mathcal{T}$ is not the discrete topology.
So we have $\mathcal{T}$ is a topology finer than the usual topology on $X$ and strictly coarser than the the discrete topology. I need to find out if $\mathcal{T}$ is different from the usual topology ( I suspect it is) and , in that case, find a basis of $\mathcal{T}$. But I don't know how to construct an open set for this topology that is not a usual open set.

A:

Yes, the topology is different. This answer gives an explicit open set which is not standard-open.
The topology itself is always a base. The "open crosses" themselves aren't open, so only function as a network, not a base. 

