# Finite intersection property

In general topology, a branch of mathematics, a collection A of subsets of a set X is said to have the finite intersection property if the intersection over any finite subcollection of A is nonempty.

A centered system of sets is a collection of sets with the finite intersection property.

## Definition

Let $X$ be a set with $A=\{A_i\}_{i\in I}$ a family of subsets of $X$. Then the collection $A$ has the finite intersection property (fip), if any finite subcollection $J\subset I$ has non-empty intersection $\bigcap_{i\in J} A_i.$

## Discussion

Clearly the empty set cannot belong to any collection with the f.i.p. 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, Xi 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 Xi 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 satisfying the finite intersection property has nonempty intersection itself.[1] 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 compact Hausdorff space that satisfies the property that no one-point set is open. If X has more than one point, then X is uncountable.

Before proving this, we give some examples:

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 given the discrete topology shows.

Proof of theorem:

Let X be a compact Hausdorff space. We will show that if U is a nonempty, open subset of X and if x is a point of X, then there is a neighbourhood V contained in U whose closure doesn’t contain x (x may or may not be in U). First of all, 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 ƒ is a bijective function from Z (the positive integers) to X. Denote the points of the image of Z under ƒ as {x1, x2, ……}. Let X be the first open set and choose a neighbourhood U1 contained in X whose closure doesn’t contain x1. Secondly, choose a neighbourhood U2 contained in U1 whose closure doesn’t contain x2. Continue this process whereby choosing a neighbourhood Un+1 contained in Un whose closure doesn’t contain xn+1. Note that the collection {Ui} for i in the positive integers 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 xi can belong to this intersection because xi doesn’t belong to the closure of Ui. This means that x is not equal to xi for all i and ƒ is not surjective; a contradiction. Therefore, X is uncountable.

Corollary

Every closed interval [ab] (a < b) is uncountable. Therefore, the set of real numbers is uncountable.

Corollary

Every locally compact Hausdorff space that is also perfect is uncountable.

Proof

Suppose X is a locally compact Hausdorff space that is perfect and compact. Then it immediately follows that X is uncountable (from the theorem). If X is a locally compact Hausdorff, perfect space that is not compact, then the one-point compactification of X is a compact Hausdorff space that is also perfect. It follows that the one point compactification of X is uncountable. Therefore X is uncountable (deleting a point from an uncountable set still retains the uncountability of that set).

## Examples

A filter has the finite intersection property by definition.

## Theorems

Let $X \neq \emptyset$, $F \subseteq 2^X$, F having the finite intersection property. Then there exists an $F^\prime$ ultrafilter (in $2^X$) such that $F \subseteq F^\prime$. See details and proof in Csirmaz & Hajnal (1994).[2] 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.

## References

1. ^
2. ^ Csirmaz, László; Hajnal, András (1994), Matematikai logika (In Hungarian), Budapest: Eötvös Loránd University.