# Show that any finite field has exactly $2^n$ sets using atoms of the field expressed in terms of sets in the field

Let $$(\Omega,\mathcal{F})$$ be a measure space. A set $$A$$ in $$\mathcal{F}$$ is said to be an atom if there is no non-empty proper subset of $$A$$ which is in $$\mathcal{F}$$. If $$\mathcal{F}$$ is finite, describe its atoms in terms of interesection and complements of sets from $$\mathcal{F}$$. Use this to show that any finite field has exactly $$2^n$$ sets.

If $$A$$ is an atom in $$\mathcal{F}$$, then $$\forall B\subset A$$ (proper), we have either $$B=\emptyset$$ or $$B\notin\mathcal{F}$$. Now, for all $$B_i\in\mathcal{F}$$, we have
(i) $$A\subseteq B_i$$
(ii) $$A\subseteq B_i^c$$
(iii) $$C=B_i\cap A\subset A$$, which is an impossibility.
Therefore, $$A=\cap_{i=1}^n C_i$$ where $$C_i=\begin{cases}B_i&\text{if, }A\subseteq B_i\\B_i^c&\text{if, }A\subseteq B_i\end{cases}$$ where, $$\mathcal{F}=n$$.

Is this correct? How do I use this to show the last part?

The following holds for any field $$\mathcal{F}$$ (if it's not apparent to you, you should show it's true, using only simple manipulation of sets in a field):

Let $$\mathcal{A} \subset \mathcal{F}$$ denote the subset of atoms of $$\mathcal{F}$$.
For each $$B \in \mathcal{F}$$, we can define $$\mathcal{A}_B = \{A \in \mathcal{A} : A \subseteq B\} \subseteq \mathcal{A},$$ and for each finite $$\mathcal{B} \subseteq \mathcal{A}$$ we have $$\bigcup \mathcal{B} \in \mathcal{F}$$.

Then, for any finite $$\mathcal{B} \subseteq \mathcal{A}$$, we have $$\mathcal{B} = \mathcal{A}_{\bigcup \mathcal{B}}$$ and for any $$B \in \mathcal{F}$$, we have $$\bigcup \mathcal{A}_B \subseteq B$$.

That's as far as we can go with general fields. For finite fields, however, the restriction to finite subsets of $$\mathcal{A}$$ becomes trivial, and we can show that $$\bigcup \mathcal{A}_B = B$$, which will show that $$|\mathcal{F}| = |\mathcal{P}(\mathcal{A})| = 2^{|\mathcal{A}|}$$.

To do this, we may use what you've already shown, that "All atoms have the form $$\bigcap_{i=1}^m C_i$$" where $$C_i$$ is either $$B_i$$ or its complement, along with a fact that you may have missed that "All sets of the form $$\bigcap_{i=1}^m C_i$$ are either atoms or empty".

Specifically, suppose $$B \in \mathcal{F}$$ is given by $$B=B_j$$ for some $$1 \leq j\leq m$$. Then the union of all the sets $$\bigcap_{i=1}^m C_i$$ such that $$C_j = B_j$$ is the union of atoms and empty sets and it equals $$B_j$$.

Do you mean $$2^n$$ subsets? I would say: define $$\mathcal{F} : = \mathcal{P} (X)$$, then you have $$A$$ atom iff $$A = \{a\}, a \in X$$ of cardinality $$n$$.

With your previous analysis, instead of choosing $$B_i$$ or $$B_i^c$$, chose only the $$B_i$$ that verify $$\{a\} \subseteq B_i$$, from that, you define $$f(B) \in \{0,1\}^{X}$$ by $$f(B)_a = 1$$ iff $$\{a \} \subseteq B_i$$ for $$a \in X$$ and $$B \in \mathcal{F}$$. Show that it is bijection (with your previous work) then you have $$\sharp F = 2^{\sharp X} = 2^n$$.

• Your first paragraph is odd. Just to check: Are you answering a question about a field of sets or about the algebraic structure using the same name? Commented Jun 19, 2023 at 20:57
• @BrianMoehring I get the misunderstanding (from the language barrier) I don't know why English chose the same names for the algebraic structure and the measurement theory structure. In French it is called a "Tribe" which means tribe.. I will edit my post thanks for the comment (I originally answered thinking he wanted to compute $\sharp \mathcal{P}(k)$ with $k$ a finite field of cardinality $n$ which seemed odd to me...) Commented Jun 19, 2023 at 21:49
• Note that my concern was about the entire first paragraph (even the part still there). I would interpret the underlying problem as showing that every finite field of sets (a finite tribe/clan in the French school) is order-isomorphic to $\mathcal{P}(X)$ for a finite set $X$, so assuming $\mathcal{F} := \mathcal{P}(X)$ at best leaves off the main part of the argument showing how we may simplify to that case. Commented Jun 19, 2023 at 23:47