# Let X be a set. How many $\sigma$-algebras of subsets of X contain exactly $5$ elements?

One of the questions on a past final for a measure theory course I'm taking is "Let X be a set. How many $$\sigma$$-algebras of subsets of X contain exactly $$5$$ elements?". Would the answer to this question just be $$infinity$$ since we don't have any information about the set X? Apologies if this question may seem obvious but I find measure theory extremely difficult to wrap my head around.

• Can you come up with an example of a set $X$ and a $\sigma$-algebra on $X$ that has 5 elements? – Alonso Delfín Jul 31 '20 at 22:57
• By any chance is it impossible to have a $\sigma$-algebra on X with 5 elements since we must always have a set and its compliment in that $\sigma$-algebra which would thus mean there is always an even number of elements? If I'm way off I apologize! – Jack Jul 31 '20 at 23:03
• Yes! In general, it can be shown that any $\sigma$-algebra with a finite number of elements must have $2^n$ elements for some $n$. If you are interested, I can add a sketch of a proof for this fact as an answer. – Alonso Delfín Jul 31 '20 at 23:05
• @AlonsoDelfín There is one, here: math.stackexchange.com/questions/181555/… – uniquesolution Jul 31 '20 at 23:07
• Thank you so much for your help! A sketch would be amazing, thank you. – Jack Jul 31 '20 at 23:07

Theorem. If $$\mathcal{F}$$ is a $$\sigma$$-algebra on $$X$$ that has a finite number of elements, then $$\mathrm{card}(X)=2^n$$ for some $$n$$.
Sketch of proof. Assume w.l.o.g. that $$X \neq \varnothing$$. For each $$x \in X$$ let $$\mathcal{F}_x:=\{ E \in \mathcal{F} : x \in E\}$$ Clearly each $$\mathcal{F}_x \neq \varnothing$$. We say $$x \sim y$$ in $$X$$ if $$\mathcal{F}_x = \mathcal{F}_y$$. This gives an equivalence relation on $$X$$. Some useful facts that you can check are:
• that the equivalence classes are actually given by $$[x]=\bigcap_{E \in \mathcal{F}_x}E \in \mathcal{F}$$
• that for each $$E \in \mathcal{F}$$ $$E=\bigsqcup_{x \in E} [x]$$ where $$\bigsqcup_{x \in E}$$ means that the union is disjoint (i.e. if $$x, y \in E$$ are such that $$x \neq y$$, then $$[x] \neq [y]$$).
Let $$\mathscr{A}$$ be the set whose elements are distinct equivalence classes of $$\sim$$ and put $$n := \mathrm{card}(\mathscr{A})$$. If you want a clear picture, this means that after selecting $$n$$ representatives you have $$\mathscr{A}=\{[x_1], \ldots, [x_n]\}$$ Let $$\mathcal{P}(\mathscr{A})$$ be the power set of $$\mathscr{A}$$. From basic set theory we know that $$\mathrm{card}(\mathcal{P}(\mathscr{A}))=2^n$$. Finally, define a map $$\varphi: \mathcal{P}(\mathscr{A}) \to \mathcal{F}$$ by letting $$\varphi(\mathscr{B}):= \bigsqcup_{[x] \in \mathscr{B}} [x]$$ for any $$\mathscr{B} \subseteq \mathscr{A}$$ with $$\mathscr{B}\neq \varnothing$$ and $$\varphi(\varnothing):=\varnothing$$. It is not hard at this point to check that $$\varphi$$ is a bijection and therefore that $$\mathrm{card}(\mathcal{F})=2^n$$. $$\blacksquare$$