In general topology, a branch of mathematics, a collection A of subsets of a setX is said to have the finite intersection property if the intersection over any finite subcollection of A is non-empty. It has the strong finite intersection property if the intersection over any finite subcollection of A is infinite. A centered system of sets is a collection of sets with the finite intersection property.
Definition
Let be a set and let a family of subsets of indexed by an arbitrary set. Then the collection has the finite intersection property, if any finite subcollection has non-empty intersection, that is is a non-empty subset of
Discussion
Clearly the empty set cannot belong to any collection with the finite intersection property. The condition is trivially satisfied if the intersection over the entire collection is non-empty, and it is also trivially satisfied if the collection is nested, meaning that the collection is totally ordered by inclusion, e.g. the nested sequence of intervals. These are not the only possibilities however. For example, if X = and for each positive integeri, Xi is the set of elements of X having a decimal expansion with digit 0 in the ith decimal place, then any finite intersection is non-empty, but the intersection of all Xi for i ≥ 1 is empty, since no element of has all zero digits. The finite intersection property is useful in formulating an alternative definition of compactness: a space is compactif and only if every collection of closed sets having the finite intersection property has non-empty intersection. This formulation of compactness is used in some proofs of Tychonoff's theorem and the uncountability of the real numbers.
Applications
Theorem. Let X be a non-empty compact Hausdorff space that satisfies the property that no one-point set is open. Then X is uncountable. Proof. We will show that if U ⊆ X is non-empty and open, and if x is a point of X, then there is a neighbourhoodV ⊂ U whose closure does not contain x. Choose y in U different from x. Then by the Hausdorff condition, choose disjoint neighbourhoods W and K of x and y respectively. Then K ∩ U will be a neighbourhood of y contained in U whose closure doesn't contain x as desired. Now suppose f: N → X is a bijection, and let denote the image off. Let X be the first open set and choose a neighbourhood U1 ⊂ X whose closure does not contain x1. Secondly, choose a neighbourhood U2 ⊂ U1 whose closure does not contain x2. Continue this process whereby choosing a neighbourhood Un+1 ⊂ Un whose closure does not contain xn+1. Then the collection satisfies the finite intersection property and hence the intersection of their closures is non-empty by the compactness of X. Therefore, there is a point x in this intersection. No xi can belong to this intersection because xi does not belong to the closure of Ui. This means that x is not equal toxi for all i and f is not surjective; a contradiction. Therefore, X is uncountable. All the conditions in the statement of the theorem are necessary: 1. We cannot eliminate the Hausdorff condition; a countable set with the indiscrete topology is compact, has more than one point, and satisfies the property that no one point sets are open, but is not uncountable. 2. We cannot eliminate the compactness condition, as the set of rational numbers shows. 3. We cannot eliminate the condition that one point sets cannot be open, as any finite space with the discrete topology shows. Corollary. Every closed interval with a < b is uncountable. Therefore, R is uncountable. Corollary. Every perfect, locally compact Hausdorff space is uncountable. Proof. Let X be a perfect, compact, Hausdorff space, then the theorem immediately implies that X is uncountable. If X is a perfect, locally compact Hausdorff space that is not compact, then the one-point compactification of X is a perfect, compact Hausdorff space. Therefore, the one point compactification of X is uncountable. Since removing a point from an uncountable set still leaves an uncountable set, X is uncountable as well.
Examples
A proper filter on a set has the finite intersection property.
Theorems
Let X be non-empty, F ⊆ 2X, F having the finite intersection property. Then there exists an Uultrafilter such that F ⊆ U. See details and proof in. This result is known as the ultrafilter lemma.
Variants
A family of setsA has the strong finite intersection property, if every finite subfamily of A has infinite intersection.
