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From Wikipedia's article for filter on a set:

$P(S)$ is the power set of a set $S$. Given a subset $T$ of $P(S)$, we can ask whether there exists a smallest filter $F$ containing $T$.

I was wondering why

Such a filter exists if and only if the finite intersection of subsets of $T$ is non-empty.

and why

$F$ can be constructed by taking all finite intersections of $T$ which is then filter base for $F$.

If I am correct the filter base constructed as the intermediate is also the smallest filter base containing $T$, right?

Thanks and regards!

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No, the filter base is not necessarily "smallest". For example, if $T$ had exactly three elements, the filter base constructed would consist of the sets of $T$, the pairwise intersections, and the intersection of all three elements of $T$. But you could also generate the same filter by just considering the intersection of all three elements; throw in the three elements to get a filter base that contains $T$ but is smaller than the constructed one. – Arturo Magidin Jan 30 '12 at 3:50
up vote 3 down vote accepted

If there is a finite subset of $T$ with empty intersection, then any filter that contains $T$ would necessarily contain the empty set (since filters are closed under finite intersections), and this is impossible. So the given condition is certainly necessary.

To see that the given condition is sufficient, consider the collection $U$ of all subsets of $S$ that contain a finite intersection of elements of $T$ (the empty intersection being defined to be $S$ itself). I claim that this is a filter on $S$.

Certainly, $U$ contains $S$, since $S$ contains every element of $T$. Since no finite intersection of elements of $T$ is empty, the empty set does not contain any finite intersection of elements of $T$, so $\varnothing\notin U$. If $A$ contains $T_1\cap\cdots \cap T_n$ and $B$ contains $R_1\cap\cdots\cap R_m$, where $T_i,R_j\in T$, then $A\cap B$ contains $(T_1\cap\cdots\cap T_n)\cap(R_1\cap\cdots\cap R_m)$, which is a finite intersection of elements of $T$; hence $A\cap B\in U$. And if $A\in U$ then $A$ contains a finite intersection of elements of elements of $T$, and so does any $B\subseteq S$ that has $A\subseteq B$. So if $A\in U$ and $A\subseteq B$, then $B\in U$. Thus, $U$ is a filter on $S$, as claimed. Moreover, $S$ contains $T$.

Finally, I claim that if $\{U_i\}$ is a nonempty family of filters on $S$, each of which contains $T$, then $\bigcap U_i$ is a filter on $S$ that contains $T$. Indeed, let $U$ denote the intersection. Since $S\in U_i$ for all $i$, then $S\in U$. Since $\varnothing\notin U_i$ for each $i$, $\varnothing\notin U$. If $A,B\in U$, then $A,B\in U_i$ for each $i$, so $A\cap B\in U_i$ for each $i$, hence $A\cap B\in U$. If $A\in U$ and $A\subseteq B$, then $A\in U_i$ for each $i$, hence $B\in U_i$ for each $i$, hence $B\in U$. And finally, since $T\subseteq U_i$ for each $i$, then $T\subseteq \bigcap U_i = U$.

So now simply consider $$\bigcap \{ U \mid U\text{is a filter on }S\text{ that contains }T\}.$$ We know the family is nonempty, so the intersection makes sense; it is a filter that contains $T$; and by construction it is the smallest filter on $S$ that contains $T$. Thus, the condition is sufficient.

(In fact, the specific example that was constructed above is the smallest filter that contains $T$, since any filter that contains $T$ must contain every set that contains all finite intersections of elements of $T$.)

You'll note that the one example I constructed was constructed as "the filter of all sets that contain some finite intersection of elements of $T$". That means the filter whose base are the "finite intersection of elements of $T$".

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If $\mathcal F$ is a filter containing $T$ then $A_1,...,A_n\in T$ implies $A_1\cap ...\cap A_n\in \mathcal F$ by $(3)$ and the intersection is non-empty by $(2)$. On the other hand, if the finite intersection of subsets of $T$ is non-empty consider the set $\{A\subseteq S: \exists A_1,...,A_n\in T\ \text{and}\ (A_1\cap...\cap A_n\subseteq A\}$. It is easy to see that it is a filter containing $T$.

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