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.
 A: The answer is none.
The reason is the following theorem:
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 $n$.
I'll give a sketch of a proof and I'll leave it to you to fill in all the missing details.
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 $\mathrm{card}(\mathcal{F})=2^n$. $\blacksquare$
