Cardinality of Infinite Sigma-Algebra is at Least Cardinality of Continuum

Theorem
Let $\mathcal M$ be an infinite $\sigma$-algebra on a set $X$.

Then $\mathcal M$ is uncountable, and has cardinality:


 * $\operatorname{card}\left({\mathcal M }\right) \ge \mathfrak c$

where $\mathfrak c$ is the cardinality of the continuum.

Proof
We first show that $X$ is infinite.

By the definition of a $\sigma$-algebra, $\mathcal M$ is a subset of $\mathcal P(X)$.

Were $X$ finite, by Cardinality of Power Set of Finite Set, the cardinality of $\mathcal M$ would be at most $2^{\operatorname{card}(X)}$

As $2^{\operatorname{card}(X)}$ is finite if $X$ is finite, $X$ must be infinite by the assumption that $\mathcal M$ is infinite.

By the definition of $\sigma$-algebra, $X \in \mathcal M$.

Also, by Sigma-Algebra Contains Empty Set, $\varnothing \in \mathcal M$.

Construct a countable collection of sets $\left\langle {F_1, F_2, F_3, \ldots  } \right\rangle_{\N}$ as follows:


 * $F_1 = \varnothing$


 * $F_2 = X$

We can continue this construction using the axiom of choice:


 * $F_3 = \text{ any set in } \mathcal M \setminus \left \{ { \varnothing, X } \right \}$


 * $\ \ \vdots$


 * $F_n = \text{ any set in } \mathcal M \setminus \left \{ {F_1, F_2, \ldots, F_{n-1} } \right \}$


 * $\ \ \vdots$

Consider an arbitrary $f \in \bigcup_{k \mathop \in \N} F_k$

Then $f \in F_k$ for some $F_k$.

By the well-ordering principle, there is a smallest such $k$.

Then for any $j < k$, $f \notin F_j$

Thus the sets in $\langle F \rangle$ are disjoint.

Recall $F_2 = X$ has infinitely many elements.

Then by Relative Difference between Infinite Set and Finite Set is Infinite, $\mathcal M \setminus \left\{ {F_1, F_2, F_3, \cdots } \right\}$ is infinite.

Thus this process can continue indefinitely, choosing an arbitary set in $\mathcal M$ that hasn't already been chosen for an earlier $F_k$.

By the definition of a $\sigma$-algebra, $\displaystyle \bigsqcup_{i \mathop \in \N} F_i$ is measurable.

By the definition of an indexed family, $\langle F_i \rangle_{i \mathop \in \N}$ corresponds to a mapping $\iota: \N \hookrightarrow \bigsqcup_{i \mathop \in \N} F_i$

Such a mapping $\iota$ is injective because:


 * each $F_i$ is disjoint from every other,


 * each $F_i$ contains a distinct $x \in X$,


 * $X$ has infinitely many elements

Define:


 * $\iota^*: \mathcal P \left({\N}\right) \to \mathcal M$:


 * $\iota^*(S) = \bigsqcup_{i \mathop \in S} F_i$

That is, for every $S \subseteq \N$, $\iota^*(S)$ corresponds to a way to select a countable union of the sets in $\langle F_i \rangle$.

Because any two $F_i, F_j$ are disjoint for $i \ne j$, any two distinct ways to create a union $S \mapsto \bigsqcup_{i \mathop \in S} F_i$ will result in a different union $\iota^*(S)$.

Thus $\iota^*$ is an injection into $\mathcal M$.

Then the cardinality of $\mathcal M$ is at least $\mathcal P \left({\N}\right)$.

From Continuum equals Cardinality of Power Set of Naturals, $\R \sim \mathcal P \left({\N}\right)$.

Thus $\mathcal M$ is uncountable and $\operatorname{card}\left({\mathcal M}\right) \ge \mathfrak c$