Finite intersection property

Definition

Let X be a set with A = { A i } i ∈ I a family of subsets of X . Then the collection A has the finite intersection property (FIP), if any finite subcollection J ⊆ I has non-empty intersection ⋂ i ∈ J A i .

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 nonempty (in particular, if the collection itself is empty), and it is also trivially satisfied if the collection is nested, meaning that the collection is totally ordered by inclusion (equivalently, for any finite subcollection, a particular element of the subcollection is contained in all the other elements of the subcollection), e.g. the nested sequence of intervals (0, 1/n). These are not the only possibilities however. For example, if X = (0, 1) and for each positive integer i, X_i is the set of elements of X having a decimal expansion with digit 0 in the i'th decimal place, then any finite intersection is nonempty (just take 0 in those finitely many places and 1 in the rest), but the intersection of all X_i for i ≥ 1 is empty, since no element of (0, 1) has all zero digits.

The finite intersection property is useful in formulating an alternative definition of compactness: a space is compact if and only if every collection of closed sets having the finite intersection property has nonempty intersection. This formulation of compactness is used in some proofs of Tychonoff's theorem and the uncountability of the real numbers (see next section)

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 nonempty and open, and if x is a point of X, then there is a neighbourhood V ⊂ U whose closure doesn’t contain x (x may or may not be in U). Choose y in U different from x (if x is in U, then there must exist such a y for otherwise U would be an open one point set; if x isn’t in U, this is possible since U is nonempty). 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 {x_i : i ∈ N} denote the image of f. Let X be the first open set and choose a neighbourhood U₁ ⊂ X whose closure doesn’t contain x₁. Secondly, choose a neighbourhood U₂ ⊂ U₁ whose closure doesn’t contain x₂. Continue this process whereby choosing a neighbourhood U_n+1 ⊂ U_n whose closure doesn’t contain x_n+1. Then the collection {U_i : i ∈ N} satisfies the finite intersection property and hence the intersection of their closures is nonempty (by the compactness of X). Therefore, there is a point x in this intersection. No x_i can belong to this intersection because x_i doesn’t belong to the closure of U_i. This means that x is not equal to x_i 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 all rational numbers shows.

3. We cannot eliminate the condition that one point sets cannot be open as a finite space as the discrete topology shows.

Corollary. Every closed interval [a, b] 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 which 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 filter has the finite intersection property by definition.

Theorems

Let X be nonempty, F ⊆ 2^X, F having the finite intersection property. Then there exists an F′ ultrafilter (in 2^X) such that F ⊆ F′.

See details and proof in Csirmaz & Hajnal (1994). This result is known as ultrafilter lemma.

Variants

A family of sets A has the strong finite intersection property (sfip), if every finite subfamily of A has infinite intersection.