# 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

Let:

$\mathbb M_\infty := \set {A \in \MM : \map \Card {\MM_A} = \infty}$

where:

$\MM_A$ denotes the trace $\sigma$-algebra of $A$ in $\MM$

Let $A \in \mathbb M_\infty$.

Observe:

 $\ds \map \Card {\MM_B} + \map \Card {\MM_{A \mathop \setminus B} }$ $\ge$ $\ds \map \Card {\MM_B \cup \MM_{A \mathop \setminus B} }$ $\ds$ $\ge$ $\ds \map \Card {\MM_A}$ $\ds$ $=$ $\ds \infty$

Thus:

$\forall B \in \MM_A : B \in {\mathbb M}_\infty \lor A \setminus B \in {\mathbb M}_\infty$

In particular:

$\forall A \in {\mathbb M}_\infty : \exists B \in {\mathbb M}_\infty : B \subsetneq A$

That is:

$\O \not \in \mathbb S$

where:

$\mathbb S := \set {\set {\tuple {A, B} : B \in \mathbb M_\infty, B \subsetneq A} : A \in \mathbb M_\infty}$

By Axiom of Choice, there is a choice function:

$\tilde f : \mathbb S \to \bigcup \mathbb S$

In particular, it satisfies:

$\forall A \in \mathbb M_\infty : \map {\tilde f} {\set {\tuple {A, B} : B \in \mathbb M_\infty, B \subsetneq A} } \in \set {\tuple {A, B} : B \in \mathbb M_\infty, B \subsetneq A}$

Hence there exists a mapping:

$f : \mathbb M_\infty \to \mathbb M_\infty$

such that:

$\forall A \in \mathbb M_\infty : \map f A \subsetneq A$

More specifically, we can define:

$\map f A := \map {\pr_2} {\map {\tilde f} {\set {\tuple {A, B} : B \in \mathbb M_\infty, B \subsetneq A} } }$

where $\pr_2$ denotes the second projection of $\paren {\mathbb M_\infty}^2$.

Let:

$F_i := \map {f^i} X \setminus \map {f^{i + 1} } X$

for $i \in \N$.

Then $\sequence {F_i}_{i \mathop \in \N}$ are non-empty pairwise disjoint sets in $\MM$.

We can define an injection $\powerset \N \to \MM$ by:

$\ds N \mapsto \bigsqcup_{i \mathop \in N} F_i$

That is:

$\map \Card \MM \ge \map \Card {\powerset \N}$
$\map \Card {\powerset \N} = \mathfrak c$

Hence the result.

$\blacksquare$

## Axiom of Choice

This theorem depends on the Axiom of Choice.

Because of some of its bewilderingly paradoxical implications, the Axiom of Choice is considered in some mathematical circles to be controversial.

Most mathematicians are convinced of its truth and insist that it should nowadays be generally accepted.

However, others consider its implications so counter-intuitive and nonsensical that they adopt the philosophical position that it cannot be true.
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