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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?

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Let $\mathscr{B}_0=\{(a,b)\times(c,d):a,b,c,d\in\Bbb Q\text{ and }a<b\text{ and }c<d\}$, 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.

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