This is exercise $1.3$ in Folland's Real Analysis: Modern Techniques and Their Applications. I'd like this proofread please. The theorem is out there on math.se but I'd like this particular proof checked, kthx
Theorem: If $\mathcal M$ is an infinite $\sigma$-algebra on $X$, $\mathcal M$ has cardinality at least that of the continuum.
Proof:
1) I proved that $X$ is infinite
2) $X, \varnothing \in \mathcal M$ so I can define a sequence using the axiom of choice:
$F_1 = \varnothing, F_2 = X$,
$F_3 = \text{ any set in } \mathcal M \setminus \left \{ { \varnothing, X } \right \}; \ldots ;F_n = \text{ any set in } \mathcal M \setminus \left \{ {F_1, F_2, \ldots, F_{n-1} } \right \}$
I prove there are a countably infinite number of disjoint sets in this construction.
3) Any indexed family $\langle F_i \rangle$ so constructed is a set of measurable sets, the disjoint union of which is also measurable.
Each family such constructed corresponds to an mapping $\iota: \mathbb N \hookrightarrow \bigsqcup_{i \mathop \in \mathbb N} F_i$, by the definition of an indexed family.
I prove that each $\iota$ is an injection, using the disjointness of the sets in $\langle F \rangle$.
4) $\displaystyle \iota^\to : \mathcal P (\mathbb N) \hookrightarrow \mathcal P \left({ \bigsqcup_{i \mathop \in \mathbb N} F_i}\right)$
The induced map is an injection because an injection induces an injection on the power sets.
MISTAKE FOUND HERE in Umberto's answer
$\mathcal P \left({ \bigsqcup_{i \mathop \in \mathbb N} F_i}\right)$ is a set of measurable sets in $\mathcal M$ so $\mathcal P \left({ \bigsqcup_{i \mathop \in \mathbb N} F_i}\right) \subseteq \mathcal M$
5) Show there is an injection from $\mathbb R \hookrightarrow \mathcal M$ using $\mathcal P (\mathbb N) \sim \mathbb R$
$\blacksquare$?
Your mapping looks off. In particular, $\mathcal P(\bigsqcup_{i \in \mathbb N} F_i) \subseteq \cal M$ can't be right.
How about defining $\iota : \mathcal P(\mathbb N) \to \mathcal M$ by $$\iota(S) = \bigsqcup_{i \in S} F_i$$ and showing $\iota$ is an injection. This gives you what you need.