Finite intersection property

In general topology, a branch of mathematics, a non-empty family A of subsets of a set is said to have the finite intersection property (FIP) if the intersection over any finite subcollection of is non-empty. It has the strong finite intersection property (SFIP) if the intersection over any finite subcollection of is infinite.

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

Definition

Let be a set and let be a non-empty family of subsets of indexed by an arbitrary set . The collection has the finite intersection property (FIP) if any finite subcollection of two or more sets has non-empty intersection, that is, is a non-empty set for every non-empty finite

If is a non-empty family of sets then the following are equivalent:

  1. has the finite intersection property.
  2. The π–system generated by does not have the empty set as an element.
  3. is a filter subbase.
  4. is a subset of some prefilter.
  5. is a subset of some proper filter.

Variations

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

Discussion

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 (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), for example the nested sequence of intervals However, these are not the only possibilities. For example, if and for each positive integer is the set of elements of having a decimal expansion with digit in the th decimal place, then any finite intersection is non-empty (just take in those finitely many places and in the rest), but the intersection of all for is empty, since no element of has all zero digits.

Applications

The finite intersection property is useful in formulating an alternative definition of compactness:

Theorem: a space is compact if and only if every family of closed subsets having the finite intersection property has non-empty intersection.[1][2]

This formulation of compactness is used in some proofs of Tychonoff's theorem and the uncountability of the real numbers (see next section).

Theorem  Let be a non-empty compact Hausdorff space that satisfies the property that no one-point set is open. Then is uncountable.

Proof

We will show that if is non-empty and open, and if is a point of then there is a neighbourhood whose closure does not contain (' may or may not be in ). Choose different from (if then there must exist such a for otherwise would be an open one point set; if this is possible since is non-empty). Then by the Hausdorff condition, choose disjoint neighbourhoods and of and respectively. Then will be a neighbourhood of contained in whose closure doesn't contain as desired.

Now suppose is a bijection, and let denote the image of Let be the first open set and choose a neighbourhood whose closure does not contain Secondly, choose a neighbourhood whose closure does not contain Continue this process whereby choosing a neighbourhood whose closure does not contain Then the collection satisfies the finite intersection property and hence the intersection of their closures is non-empty by the compactness of Therefore, there is a point in this intersection. No can belong to this intersection because does not belong to the closure of This means that is not equal to for all and is not surjective; a contradiction. Therefore, is uncountable.

All the conditions in the statement of the theorem are necessary:

1. We cannot eliminate the Hausdorff condition; a countable set (with at least two points) 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 is uncountable. Therefore, is uncountable.

Corollary  Every perfect, locally compact Hausdorff space is uncountable.

Proof

Let be a perfect, compact, Hausdorff space, then the theorem immediately implies that is uncountable. If is a perfect, locally compact Hausdorff space that is not compact, then the one-point compactification of is a perfect, compact Hausdorff space. Therefore, the one point compactification of is uncountable. Since removing a point from an uncountable set still leaves an uncountable set, is uncountable as well.

Examples

A proper filter on a set has the finite intersection property. A π–system has the finite intersection property if and only if it does not have the empty set as an element.

Theorems

Let be non-empty, having the finite intersection property. Then there exists an ultrafilter (in ) such that

See details and proof in Csirmaz & Hajnal (1994).[3] This result is known as the ultrafilter lemma.

References

  1. Munkres, James (2004). Topology. New Delhi: Prentice-Hall of India. p. 169. ISBN 978-81-203-2046-8.
  2. A space is compact iff any family of closed sets having fip has non-empty intersection at PlanetMath.
  3. Csirmaz, László; Hajnal, András (1994), Matematikai logika (In Hungarian), Budapest: Eötvös Loránd University.
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