# Finite intersection property

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

## Definition

Let X{\textstyle X} be a set and A{\textstyle {\mathcal {A}}} a nonempty family of subsets of X{\textstyle X}; that is, A{\textstyle {\mathcal {A}}} is a nonempty subset of the power set of X{\textstyle X}. Then A{\textstyle {\mathcal {A}}} 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, A{\textstyle {\mathcal {A}}} has the FIP if, for any choice of a finite nonempty subset B{\textstyle {\mathcal {B}}} of A{\textstyle {\mathcal {A}}}, there must exist a point $x\in \bigcap _{B\in {\mathcal {B}}}{B}{\text{.}}$ Likewise, A{\textstyle {\mathcal {A}}} has the SFIP if, for every choice of such B{\textstyle {\mathcal {B}}}, there are infinitely many such x{\textstyle x}.
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 ${\mathcal {A}}$ is finite, then ${\mathcal {A}}$ has the finite intersection property if and only if it is fixed.

### Pairwise intersection

The finite intersection property is strictly stronger than pairwise intersection; the family $\{\{1,2\},\{2,3\},\{1,3\}\}$ has pairwise intersections, but not the FIP.
More generally, let n∈N∖{1}{\textstyle n\in \mathbb {N} \setminus \{1\}} be a positive integer greater than unity, [n]={1, …, n}{\textstyle [n]=\{1, \dots , n\}}, and A={[n]∖{j}:j∈[n]}{\textstyle {\mathcal {A}}=\{[n]\setminus \{j\}:j\in [n]\}}. Then any subset of ${\mathcal {A}}$ with fewer than n{\textstyle n} elements has nonempty intersection, but A{\textstyle {\mathcal {A}}} lacks the FIP.

### End-type constructions

If $A_{1}\supseteq A_{2}\supseteq A_{3}\cdots$ is a decreasing sequence of non-empty sets, then the family A={A$_1$, A$_2$, A$_3$, …}{\textstyle {\mathcal {A}}=\left\{A_{1}, A_{2}, A_{3}, \ldots \right\}} has the finite intersection property (and is even a π–system). If the inclusions $A_{1}\supseteq A_{2}\supseteq A_{3}\cdots$ are strict, then A{\textstyle {\mathcal {A}}} admits the strong finite intersection property as well.
More generally, any A{\textstyle {\mathcal {A}}} that is totally ordered by inclusion has the FIP.
At the same time, the kernel of A{\textstyle {\mathcal {A}}} may be empty: if A$_j$={j, j+1, j+2, …}{\textstyle A_{j}=\{j, j+1, j+2, \dots \}}, then the kernel of ${\mathcal {A}}$ is the empty set. Similarly, the family of intervals $\left\{[r,\infty ):r\in \mathbb {R} \right\}$ also has the (S)FIP, but empty kernel.

### "Generic" sets and properties

The family of all Borel subsets of $[0,1]$ with Lebesgue measure 1{\textstyle 1} has the FIP, as does the family of comeagre sets. If X{\textstyle X} is an infinite set, then the Fréchet filter (the family {X∖C:C finite}{\textstyle \{X\setminus C:C{\text{ finite}}\}}) has the FIP. All of these are free filters; they are upwards-closed and have empty infinitary intersection.
If $X=(0,1)$ and, for each positive integer $i,$ the subset $X_{i}$ is precisely all elements of $X$ having digit $0$ in the $i$th decimal place, then any finite intersection of $X_{i}$ is non-empty — just take $0$ in those finitely many places and $1$ in the rest. But the intersection of $X_{i}$ for all $i\geq 1$ is empty, since no element of $(0,1)$ has all zero digits.

### Extension of the ground set

The (strong) finite intersection property is a characteristic of the family A{\textstyle {\mathcal {A}}}, not the ground set X{\textstyle X}. If a family A{\textstyle {\mathcal {A}}} on the set X{\textstyle X} admits the (S)FIP and X⊆Y{\textstyle X\subseteq Y}, then A{\textstyle {\mathcal {A}}} is also a family on the set Y{\textstyle Y} with the FIP (resp. SFIP).

### Generated filters and topologies

If $K\subseteq X$ is a non-empty set, then the family ${\mathcal {A}}=\{S\subseteq X:K\subseteq S\}$ has the FIP; this family is called the principal filter on X{\textstyle X} generated by K{\textstyle K}. The subset ${\mathcal {B}}=\{I\subseteq \mathbb {R} :K\subseteq I{\text{ and }}I{\text{ an open interval}}\}$ has the FIP for much the same reason: the kernels contain the non-empty set K{\textstyle K}. If K{\textstyle K} is an open interval, then the set K{\textstyle K} is in fact equal to the kernels of A{\textstyle {\mathcal {A}}} or B{\textstyle {\mathcal {B}}}, and so is an element of each filter. But in general a filter's kernel need not be an element of the filter.
A proper filter on a set has the finite intersection property. Every neighbourhood subbasis at a point in a topological space has the FIP, and the same is true of every neighbourhood basis and every neighbourhood filter at a point (because each is, in particular, also a neighbourhood subbasis).

## Relationship to π-systems and filters

A π–system is a non-empty family of sets that is closed under finite intersections. The set $\pi ({\mathcal {A}})=\left\{A_{1}\cap \cdots \cap A_{n}:1\leq n<\infty {\text{ and }}A_{1},\ldots ,A_{n}\in {\mathcal {A}}\right\}$of all finite intersections of one or more sets from ${\mathcal {A}}$ is called the π–system generated by A{\textstyle {\mathcal {A}}}, because it is the smallest π–system having A{\textstyle {\mathcal {A}}} as a subset.
The upward closure of $\pi ({\mathcal {A}})$ in X{\textstyle X} is the set $\pi ({\mathcal {A}})^{\uparrow X}=\left\{S\subseteq X:P\subseteq S{\text{ for some }}P\in \pi ({\mathcal {A}})\right\}{\text{.}}$For any family A{\textstyle {\mathcal {A}}}, the finite intersection property is equivalent to any of the following:
* The π–system generated by ${\mathcal {A}}$ does not have the empty set as an element; that is, $\varnothing \notin \pi ({\mathcal {A}}).$
* The set $\pi ({\mathcal {A}})$ has the finite intersection property.
* The set $\pi ({\mathcal {A}})$ is a (proper) prefilter.
* The family ${\mathcal {A}}$ is a subset of some (proper) prefilter.
* The upward closure $\pi ({\mathcal {A}})^{\uparrow X}$ is a (proper) filter on $X$. In this case, $\pi ({\mathcal {A}})^{\uparrow X}$ is called the filter on $X$ generated by ${\mathcal {A}}$, because it is the minimal (with respect to $\,\subseteq \,$) filter on $X$ that contains ${\mathcal {A}}$ as a subset.
* ${\mathcal {A}}$ is a subset of some (proper) filter.

## Applications
### Compactness

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.
This formulation of compactness is used in some proofs of Tychonoff's theorem.

### Uncountability of perfect spaces

Another common application is to prove that the real numbers are uncountable.
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.
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 $U\subseteq X$ is non-empty and open, and if $x$ is a point of $X,$ then there is a neighbourhood $V\subset U$ whose closure does not contain $x$ ($x$' may or may not be in $U$). Choose $y\in U$ different from $x$ (if $x\in U$ then there must exist such a $y$ for otherwise $U$ would be an open one point set; if $x\notin U,$ this is possible since $U$ is non-empty). Then by the Hausdorff condition, choose disjoint neighbourhoods $W$ and $K$ of $x$ and $y$ respectively. Then $K\cap U$ will be a neighbourhood of $y$ contained in $U$ whose closure doesn't contain $x$ as desired.
Now suppose $f:\mathbb {N} \to X$ is a bijection, and let $\left\{x_{i}:i\in \mathbb {N} \right\}$ denote the image of $f.$ Let $X$ be the first open set and choose a neighbourhood $U_{1}\subset X$ whose closure does not contain $x_{1}.$ Secondly, choose a neighbourhood $U_{2}\subset U_{1}$ whose closure does not contain $x_{2}.$ Continue this process whereby choosing a neighbourhood $U_{n+1}\subset U_{n}$ whose closure does not contain $x_{n+1}.$ Then the collection $\left\{U_{i}:i\in \mathbb {N} \right\}$ 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 $x_{i}$ can belong to this intersection because $x_{i}$ does not 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.
Corollary—Every closed interval $[a,b]$ with $a<b$ is uncountable. Therefore, $\mathbb {R}$ is uncountable.
Corollary—Every perfect, locally compact Hausdorff space is uncountable.
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.

### Ultrafilters

Let $X$ be non-empty, $F\subseteq 2^{X}.$ $F$ having the finite intersection property. Then there exists an $U$ ultrafilter (in $2^{X}$) such that $F\subseteq U.$ This result is known as the ultrafilter lemma.