Does there exist sigma algebra whose cardinality is countably infinite? If yes tell me some examples. If not how to show every infinite sigma algebra is uncountable?
1 Answers
No countably infinite $\sigma$-algebras exist and here's a proof.
Let $\mathcal{A}$ be a $\sigma$-algebra with infinitely many members and $\mathcal{S}$ a subset of $\mathcal{A}$ with countably infinitely many members such that $\mathcal{S}$ covers the underlying space $\Omega$. Such a collection can always be found simply by adding the set $\{\Omega\}$ to any countably infinite collection of members of $\mathcal{A}$. Next we define a function $f \colon \Omega \to \mathcal{A}$ where $f(x)$ is the intersection of all members of $\mathcal{S}$ of which $x$ is a member. Since $\mathcal{S}$ is countably infinite, the intersection is countably infinite and thus $f(x)$ really does belong to $\mathcal{A}$. Clearly the function $f$ forms a partition of $\Omega$ and since every member $S$ of $\mathcal{S}$ is a union of the sets $f(x)$, $x \in S$, the partition has to be infinite. If the partition is uncountable, we are done, and if it is countably infinite, arbitrary unions of its members are all in $\mathcal{A}$, but there are uncountably many such unions.
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very nice argument. thanks. – GA316 Apr 15 '13 at 13:25
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I don't understand your definition of function $f$. – Mariana Nov 22 '21 at 18:02