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Let $F \subseteq \mathcal{P}(X)$ be a family of subsets of $X$. Then, it's easy to prove that there exists a minimal algebra of sets $\mathcal{A}$ that contains $F$, by putting all algebras on $X$ containing $F$ into a set and then taking its intersection. Also, if this $F$ is countable, then $\mathcal{A}$ is also countable, which is cleverly explained in Algebra generated by countable family of sets is countable?.

However, I'm having trouble proving the following property: if $|F| = n$, then $|\mathcal{A}| \leq 2^{2^n}$. I've tried "constructing" the algebra by taking all finite intersections of $F$ and adding them to $F$, then closing that set under complements, etc. etc, but it all gets too complicated after the second step. Is there a smart way to go about proving this inequality?

Matija Sreckovic
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1 Answers1

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Recall that an algebra of sets is a collection $\mathcal{A}$ of subsets of $X$ closed under finite intersections, finite unions, and complements. Suppose now that $F=\{X_1,\ldots,X_n\}$ is a collection of $n$ subsets of $X$.

The minimal algebra $\mathcal{A}$ of subsets $X$ containing $F$ is obtained by closing $X_1,\ldots,X_n$ under these operations. The atoms of such algebra are the non-empty subsets of the form $$\displaystyle{Y_\sigma=\bigcap_{i=1}^n X_i^{\sigma(i)}},$$ where $\sigma$ is a function $\sigma:\{1,\ldots,n\}\to\{0,1\}$ and we denote $X_i^0:=X_i, X_i^1:=X_i^c=X\setminus X_i$.

So, in the best case scenario, the algebra $\mathcal{A}$ has $2^n$ different elements (as many as functions from $\{1,\ldots,n\}$ to $\{0,1\}$). Now, each set in an atomic algebra can be written as a union of atoms, i.e., if $Y_1,\ldots,Y_k$ are the atoms, then the elements in $\mathcal{A}$ have the form $$\bigcup_{j=1}^kY_j^{\tau}$$ where $\tau$ is a function from $\{1,\ldots,k\}$ to $\{0,1\}$.

We have then that $k\leq 2^n$ (there are at most $2^n$ atoms), and so the number of distinct sets in $\mathcal{A}$ is at most $2^k=2^{2^n}$.

Darío G
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