This is my attempt at an exercise from Folland's Real Analysis. Could someone evaluate it (particularly the second claim)?

Let $\mathcal M$ be an infinite $\sigma$-algebra.

Claim: $\mathcal M$ contains an infinite sequence of disjoint sets.

Proof: Since $\mathcal M$ is infinite, it must contain a sequence $\left\{E_j\right\}_{j=1}^\infty$ such that $E_n\neq E_m$ whenever $n\neq m$. Let $F_j=E_j\setminus\bigcup_{k=1}^{j-1}E_k$. Then $\left\{F_j\right\}_{j=1}^\infty$ is an infinite sequence in $\mathcal M$ of disjoint sets.

Claim: $\text{card}\left(\mathcal M\right)\geq\mathfrak c$.

Proof: Let $C=\left\{0,1\right\}^\mathbb N$. Then $\text{card}\left(C\right)=\mathfrak c$. Define $f:C\to\mathcal M$ by $c\mapsto\bigcup\left\{F_j:c_j=1,j\in\mathbb N\right\}$. Then $f$ is well-defined and injective: suppose otherwise that $c,d\in C$ with $f(c)=f(d)$ and $c\neq d$. Let $k\in\mathbb N$ be such that $c_k\neq d_k$. Assume, without loss of generality, that $c_k=1$. Then, since the $F_j$'s are disjoint, there is an $x\in F_k\subset f(c)$ such that $x\notin f(d)$, contradicting the assumption that $f(c)\subset f(d)$.

  • $\begingroup$ just take $A_i=\varnothing$ $\endgroup$ – Jorge Fernández Hidalgo Jun 15 '17 at 4:22
  • 4
    $\begingroup$ You need that $\mathcal{M}$ has an infinite sequence of disjoint nonempty sets. If your $(E_j)$ is a decreasing sequence, then the $(F_j)$ are all empty. $\endgroup$ – Lord Shark the Unknown Jun 15 '17 at 4:23

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