## You are here

Homefinite fields of sets

## Primary tabs

# finite fields of sets

If $S$ is a finite set then any field of subsets of $S$ (see “field of sets” in the entry on rings of sets) can be described as the set of unions of subsets of a partition of $S$.

Note that, if $P$ is a partition of $S$ and $A,B\subset P$, we have

$\displaystyle\overline{\bigcup A}$ | $\displaystyle=$ | $\displaystyle\bigcup(P\setminus A)$ | ||

$\displaystyle(\bigcup A)\cup(\bigcup B)$ | $\displaystyle=$ | $\displaystyle\bigcup(A\cup B)$ | ||

$\displaystyle(\bigcup A)\cap(\bigcup B)$ | $\displaystyle=$ | $\displaystyle\bigcup(A\cap B)$ |

so $\{\bigcup X\mid X\subset P\}$ is a field of sets.

Now assume that $\mathcal{F}$ is a field of subsets of a finite set $S$. Let us define the set of “prime elements” of $\mathcal{F}$ as follows:

$P=\{X\in(\mathcal{F}\setminus{\varnothing})\mid(Y\subset X)\wedge(Y\in\mathcal% {F})\Rightarrow(Y=\varnothing\vee Y=X)\}$ |

The choice of terminology “prime element” is meant to be a suggestive mnemonic of how the only divisors of a prime number are 1 and the number itself.

We claim that $P$ is a partition. To justify this claim, we need to show that elements of $P$ are pairwise disjoint and that $\bigcup P=S$.

Suppose that $A$ and $B$ are prime elements. Since, by definition, $A\in\mathcal{F}$ and $B\in\mathcal{F}$ and $\mathcal{F}$ is a field of sets, $A\cap B\in\mathcal{F}$. Since $A\cap B\subset A$, we must either have $A\cap B=\varnothing$ or $A\cap B=A$. In the former case, $A$ and $B$ are disjoint, whilst in the latter case $A=B$.

Suppose that $x$ is any element of $S$. Then we claim that the set $X$ defined as

$X=\bigcup\{Y\in\mathcal{F}\mid x\in Y\}$ |

is a prime element of $\mathcal{F}$. To begin, note that, since $\mathcal{F}$ is finite, a forteriori any subset of $\mathcal{F}$ is finite and, since fields of sets are assumed to be closed under intersection, it follows that the intersection of a susbet of $\mathcal{F}$ is an element of $\mathcal{F}$, in particular $X\in\mathcal{F}$.

Suppose that $Z\subset X$ and $Z\in\mathcal{F}$. If $x\notin Z$, then $x\in\overline{Z}$. Since $\mathcal{F}$ is a field of sets, $\overline{Z}\in\mathcal{F}$. Hence, by the construction of $X$, it is the case that $X\subset\overline{Z}$, hence $X\cap Z=\varnothing$. Together with $Z\subset X$, this implies $Z=\varnothing$. If $x\in Z$, then, by construction, $X\subset Z$, which implies $X=Z$.

Thus, we see that $X$ is a prime set. Since $x$ was arbitrarily chosen, this means that every element of $S$ is contained in a prime element of $\mathcal{F}$, so the union of all prime elements is $S$ itself. Together with the previously shown fact that prime elements are pairwise disjoint, this shows that the prime elements for a partition of $S$.

Let $A$ be an arbitrary element of $\mathcal{F}$. Since $P\subset\mathcal{F}$, it is the case that $(\forall X\in P)A\cap X\in\mathcal{F}$. Since $P$ is a partition of $S$,

$A=\bigcup\{A\cap X\mid X\in P\}$ |

so every element of $\mathcal{F}$ can be expressed as a union of elements of $P$.

## Mathematics Subject Classification

28A05*no label found*03E20

*no label found*

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff
- Corrections