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Proof

(related to Proposition: Filter Base)

• By hypothesis, $(X,\mathcal O)$ is a "topological space":https://www.bookofproofs.org/branches/topological-spce-topology/ and $B$ is a non-empty set of subsets $O\subseteq X$ that does not contain the empty set $\emptyset.$
• Let $F$ be a set of all subsets of $X$ that contain some element of $B.$

"$\Rightarrow$"

• Assume $F$ is a filter.
• Since $F$ is a filter, it contains the intersection of any two of its elements.
• By hypothesis, $F$ contains all subsets of $X$ containing some elements of $B$.
• Thus, the intersection of any two sets of $B$ contains a set in $B$ as an element.

"$\Leftarrow$"

• Assume, the intersection of any two sets of $B$ contains a set in $B.$
• Since $F$ contains all of the subsets of $X$ containing an element of $B$, $F$ contains the intersection of any two of its elements.
• Moreover, since $B$ does not contain the empty set, so does not $F.$
• Finally, all supersets of the elements of $B$ are contained in $F$
• It follows that $F$ is a filter.
  ∎

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References

Bibliography

1. Steen, L.A.;Seebach J.A.Jr.: "Counterexamples in Topology", Dover Publications, Inc, 1970
2. Jänich, Klaus: "Topologie", Springer, 2001, 7th Edition