# Shrinking basis of topological space

All rectangles consists a basis $$\mathcal{B}$$ for standard topology $$\mathcal{T}$$ on $$R^2$$.

If remove one rectangle from $$\mathcal{B}$$, it will still be a basis for $$\mathcal{T}$$, just like the All-Pies generating the same topology as All-rectangles thing.

It seems that removing coutable element will keep the generated topology.

Will every uncountable-removing not keep the generated topology?

Is there some theory about the general problem about removing elements from a basis of a topology which may be related to the cardinal number of the topology or the unerlying set?

Let $$\mathscr{B}_0=\{(a,b)\times(c,d):a,b,c,d\in\Bbb Q\text{ and }a, the set of open rectangles in $$\Bbb R^2$$ that are Cartesian products open intervals in $$\Bbb R$$ with rational endpoints; $$\mathscr{B}_0$$ is a base for $$\mathscr{T}$$. Clearly we get $$\mathscr{B}_0$$ by removing from $$\mathscr{B}$$ the set $$\mathscr{B}\setminus\mathscr{B}_0$$. But $$\mathscr{B}_0$$ is countable, while $$\mathscr{B}$$ is uncountable, so $$\mathscr{B}\setminus\mathscr{B}_0$$ is uncountable. Thus, it is possible to remove an uncountable subset of $$\mathscr{B}$$ and still have a base for $$\mathscr{T}$$.
There is no general statement that involves only the cardinalities of $$\mathscr{B}$$, $$\mathscr{T}$$, and the subset of $$\mathscr{B}$$ that you remove: it also depends on the specific topology on the underlying set.