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Lets let $X = P $ where $P$ is the set of all prime numbers. Lets define $ \mathcal{A} $ to be the set of all $A$'s such that each $A = \{ p \}$ where $ p \in P$ we then want to show that the Borel sigma algebra of $\mathcal{A}$ over $P$ has countably many elements.

For each set $B$ with finitely many elements in the Borel sigma. We define the map $\phi$ to take each B in the Borel sigma algebra to the product of its primes in $\mathbb{N}$ that is to say $ \phi: B \to p_1 p_2 ... p_n \in \mathbb{N}$ moreover we define $\phi : B^c \to (p_1 p_2 ... p_n -1) $ This is an injection of each set in the Borel sigma algebra to $\mathbb{N}$ this shows us that the Borel sigma algebra of $\mathcal{A}$ is countable.

For one i would like to know if this is correct and secondly there has to be a better way to state this?

Davide Giraudo
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Faust
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  • A $\sigma$ algebra over $P$ that contains every singleton ${p}$ for $p \in P$ must contain all of the continuum-many subsets of $P: it’s closed under countable unions, and every subset of $P$ is a countable union of singletons. – BrianO Sep 14 '21 at 20:23

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This is not correct: a $\sigma$-algebra has to be closed under countable unions and intersections, so you need to also consider e.g. $$\{2\}\cup\{5\}\cup\{11\}\,\cup\, ...=\{2,5,11,...\}=\{p_{2i}:i\in\mathbb{N}\}$$ the set of every other prime.

Indeed, the $\sigma$-algebra in your post will be uncountable, and there is in fact no countably infinite $\sigma$-algebra at all (see e.g. here).

Noah Schweber
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