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

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

Then $\MM$ is has cardinality at least that of the cardinality of the continuum $\mathfrak c$:


 * $\map \Card \MM \ge \mathfrak c$

Corollary
Let $\MM$ be an infinite $\sigma$-algebra on a set $X$.

Then $\MM$ is uncountable.

Proof
We first show that $X$ is infinite.

By the definition of a $\sigma$-algebra, $\MM$ is a subset of $\powerset X$.

Were $X$ finite, by Cardinality of Power Set of Finite Set, the cardinality of $\MM$ would be at most $2^{\map \Card X}$

As $2^{\map \Card X}$ is finite if $X$ is finite, $X$ must be infinite by the assumption that $\MM$ is infinite.

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

Also, by Sigma-Algebra Contains Empty Set, $\O \in \MM$.

Construct a countable collection of sets $\family {F_1, F_2, F_3, \ldots}_\N$ as follows:


 * $F_1 = \O$


 * $F_2 = X$

We can continue this construction using the axiom of choice:


 * $F_3 = \text{ any set in } \MM \setminus \set {\O, X}$


 * $\ \ \vdots$


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


 * $\ \ \vdots$

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

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

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

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

Thus the sets in $\family {F_i}$ are disjoint.

Recall $\MM$ is infinite.

Then by Relative Difference between Infinite Set and Finite Set is Infinite, $\MM \setminus \set {F_1, F_2, \cdots, F_{k - 1} }$ is infinite.

Thus this process can continue indefinitely, choosing an arbitary set in $\MM$ 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, $\family {F_i}_{i \mathop \in \N}$ 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:


 * $\iota^*: \powerset \N \to \MM$:


 * $\map {\iota^*} N = \bigsqcup_{i \mathop \in N} F_i$

That is, for every $N \subseteq \N$, $\map {\iota^*} N$ corresponds to a way to select a countable union of the sets in $\family {F_i}$.

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

Thus $\iota^*$ is an injection into $\MM$.

Then the cardinality of $\MM$ is at least $\powerset \N$.

From Continuum equals Cardinality of Power Set of Naturals, $\R \sim \powerset \N$.

Thus $\MM$ is uncountable and:
 * $\map \Card \MM \ge \mathfrak c$

Proof of Corollary
Follows from the main result, as the real numbers are uncountable.