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Let $\Omega$ be an infinite set. Show that any infinite sigma field, $\mathcal{F}$ generated by subsets of $\Omega$ is uncountable, i.e. $|\mathcal{F}|\geq 2^{\aleph_0}$. obviousy we have that $|\mathcal{F}|\leq 2^{|\Omega|}$ and $2^{\aleph_0}\leq 2^{|\Omega|}$. but am having trouble showing that$\mathcal{F}$ is uncountable as sigma algebras are made up of counbable unions etc. Any help am really stuck. Any ideas on how to define injection from $2^{\aleph_0}$ to $\mathcal{F}$, but thinking there is probably a quick way to show this with resoting to defining an injection? Thanks in advance for any assistance

user152874
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Assume that $\mathcal{F}$ is countable. In this case, arbitrary(!) intersections and unions of elements in $\mathcal{F}$ will again be in $\mathcal{F}$.

Thus, for each $x \in \Omega$, you can define $F_x := \bigcap_{A \in \mathcal{F}, x \in A} A \in \mathcal{F}$.

Show that for each $F \in \mathcal{F}$ we have $$ F = \bigcup_{x \in F} F_x.$$

Thus, $\{F_x | x \in \Omega\}$ must be an infinite set if $\mathcal{F}$ is infinite (why?).

Use this to find a subset of $\mathcal{F}$ in bijection with $2^{\Bbb{N}}$. Here it will by useful that $F_x \cap F_y = \emptyset$ if $F_x \neq F_y$.

This derives a contradiction.

EDIT: I just realized that you wanted to show $|\mathcal{F}| \geq |2^{\Bbb{N}}|$. The above proof will only show (by contradiction) that $\mathcal{F}$ is uncountable. It will not itself show $|\mathcal{F}| \geq |2^{\Bbb{N}}|$.

If you insist on showing the above estimate, you should probably look at An infinite $\sigma$-algebra contains an infinite sequence of nonempty, disjoint sets. to construct sets similar to the ones above which will yield a bijection between a subset of $\mathcal{F}$ and $2^{\Bbb{N}}$.

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PhoemueX
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  • Ok, I see that ${F_x|x\in\Omega}$ is a the partition of $\Omega$ and it must be infinite, as $\mathcal{F}$ is. Now one can form all the sets in $\mathcal{F}$ by taking all the possible (disjoint) unions of sets in ${F_x|x\in\Omega}$. But how does this means that the cardinality of $\mathcal{F}$ is equal to the cardinality of the power set of $\Omega$. Guess I am not seeing it right, can you clear it up for me. Thanks – user152874 May 23 '14 at 16:19
  • First of all, you can not show that the cardinality of $\mathcal{F}$ equals $|2^{\Omega}|$, because this is false in general. For example the Borel-$\sigma$-Algebra on $\Bbb{R}$ has cardinality $|\Bbb{R}|$ instead of $|2^{\Bbb{R}}|$. Second, let $(x_n)n$ be an infinite sequence such that the $F{x_n}$ are pairwise disjoint. Then the collection ${\bigcup_{i \in I} F_{x_i} | I \subset \Bbb{N} }$ is in bijection with $2^{\Bbb{N}}$, so $|\mathcal{F}| \geq |2^{\Bbb{N}}|$. – PhoemueX May 23 '14 at 18:13
  • Sorry that was a typo, what I meant was that $|\mathcal{F}|$ is thae same as the cardinality of the powerset of ${F_x|x\in \Omega}$ – user152874 May 23 '14 at 18:38
  • Use that $T : \Gamma \rightarrow 2^{{x_n | n \in \Bbb{N}}}, F \mapsto {x_n | x_n \in F}$ with $\Gamma = {\bigcup_{i \in I} F_{x_i} | I \subset \Bbb{N} }$ is a bijection. – PhoemueX May 23 '14 at 18:43