Cardinality of Infinite Sigma-Algebra is at Least Cardinality of Continuum/Proof 2

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:

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

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:
 * $\map \Card \MM \ge \map \Card {\powerset \N}$

But by Power Set of Natural Numbers has Cardinality of Continuum:
 * $\map \Card {\powerset \N} = \mathfrak c$

Hence the result.