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. Sets with the finite intersection property are also called centered systems and filter subbases.
The finite intersection property can be used to reformulate topological compactness in terms of closed sets; this is its most prominent application. Other applications include proving that certain perfect sets are uncountable, and the construction of ultrafilters.
Let be a set and a nonemptyfamily of subsets of ; that is, is a subset of the power set of . Then is said to have the finite intersection property if every nonempty finite subfamily has nonempty intersection; it is said to have the strong finite intersection property if that intersection is always infinite.
In symbols, has the FIP if, for any choice of a finite nonempty subset of , there must exist a point
Likewise, has the SFIP if, for every choice of such , there are infinitely many such .
In the study of filters, the common intersection of a family of sets is called a kernel, from much the same etymology as the sunflower. Families with empty kernel are called free; those with nonempty kernel, fixed.
Families of examples and non-examples
The empty set cannot belong to any collection with the finite intersection property.
A sufficient condition for the FIP intersection property is a nonempty kernel. The converse is generally false, but holds for finite families; that is, if is finite, then has the finite intersection property if and only if it is fixed.
The finite intersection property is strictly stronger than pairwise intersection; the family has pairwise intersections, but not the FIP.
More generally, let be a positive integer greater than unity, , and . Then any subset of with fewer than elements has nonempty intersection, but lacks the FIP.
If is a decreasing sequence of non-empty sets, then the family has the finite intersection property (and is even a π–system). If the inclusions are strict, then admits the strong finite intersection property as well.
If and, for each positive integer the subset is precisely all elements of having digit in the thdecimal place, then any finite intersection of is non-empty — just take in those finitely many places and in the rest. But the intersection of for all is empty, since no element of has all zero digits.
Extension of the ground set
The (strong) finite intersection property is a characteristic of the family , not the ground set . If a family on the set admits the (S)FIP and , then is also a family on the set with the FIP (resp. SFIP).
If are sets with then the family has the FIP; this family is called the principal filter on generated by . The subset has the FIP for much the same reason: the kernels contain the non-empty set . If is an open interval, then the set is in fact equal to the kernels of or , and so is an element of each filter. But in general a filter's kernel need not be an element of the filter.
All the conditions in the statement of the theorem are necessary:
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.
We cannot eliminate the compactness condition, as the set of rational numbers shows.
We cannot eliminate the condition that one point sets cannot be open, as any finite space with the discrete topology shows.
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.
Corollary — Every closed interval with is uncountable. Therefore, is uncountable.
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.
Let be non-empty, having the finite intersection property. Then there exists an ultrafilter (in ) such that This result is known as the ultrafilter lemma.
^ abA filter or prefilter on a set is proper or non-degenerate if it does not contain the empty set as an element. Like many − but not all − authors, this article will require non-degeneracy as part of the definitions of "prefilter" and "filter".
Koutras, Costas D.; Moyzes, Christos; Nomikos, Christos; Tsaprounis, Konstantinos; Zikos, Yorgos (20 October 2021). "On Weak Filters and Ultrafilters: Set Theory From (and for) Knowledge Representation". Logic Journal of the IGPL. doi:10.1093/jigpal/jzab030.