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

Proof
Let:
 * $\MM_\infty := \set { A \in \MM : \map \Card {\MM_A} = \infty}$

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

Let $A \in \MM_\infty$.

Then:
 * $\forall B \in \MM_A : B \in \MM_\infty \lor A \setminus B \in \MM_\infty$

since:
 * $\map \Card {\MM_B} + \map \Card {\MM_{A \setminus B} } \ge \map \Card {\MM_A} = \infty$

In particular:
 * $\forall A \in \MM_\infty : \exists B \in \MM_\infty : B \subsetneq A$

Thus by Axiom of Choice, there exists a mapping:
 * $f : \MM_\infty \to \MM_\infty$

such that:
 * $\forall A \in \MM_\infty : \map f A \subsetneq A$

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

for $i \in \N$.

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

Define the injection:
 * $\iota^*: \powerset \N \to \MM$:


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

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

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