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I have a little doubt about what is a generated filter. If I have a base of a topology the topology generated for this base is the set of arbitrary unions of the elements of that base. If I have a filter base $\mathcal{B}$ what does it consist the generated filter of this base?

I think that's the set of supersets of the elements of the base.

My apologies if the question is very simple but I have this doubt and I don't have someone to ask.

Thanks in advantage.

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Yep, that's right. Let $F$ be the set of supersets of elements of $\mathcal{B}$. Clearly every element of $F$ must be in the filter generated by $\mathcal{B}$, since filters are closed under taking supersets. So it suffices to show that $F$ really is a filter.

Since a superset of a superset is a superset, $F$ is closed under taking supersets. So we just have to check that $F$ is closed under pairwise intersections. Suppose $A,B\in F$. Then there are elements $C,D\in\mathcal{B}$ such that $A\supseteq C$ and $B\supseteq D$. Since $\mathcal{B}$ is a filter base, there is $E\in\mathcal{B}$ such that $E\subseteq C\cap D$. We then have $A\cap B\supseteq C\cap D\supseteq E$, so $A\cap B$ is a superset of an element of $\mathcal{B}$. Thus $A\cap B\in F$.

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