wiki gives this definition of sigma-algebra

Let X be some set, and let ${\mathcal {P}}(X)$ represent its power set. Then a subset ${\displaystyle \Sigma \subseteq {\mathcal {P}}(X)}$ is called a σ-algebra if it satisfies the following three properties:

  1. X is in Σ, and X is considered to be the universal set in the following context.
  2. Σ is closed under complementation: If A is in Σ, then so is its complement, X \ A.
  3. Σ is closed under countable unions: If $A_1, A_2, A_3, ...$ are in Σ, then so is $A = A_1 ∪ A_2 ∪ A_3 ∪ …$ .

since {X, ∅} satisfies condition (3), it follows that {X, ∅} is the smallest possible σ-algebra on X. The largest possible σ-algebra on X is $2^X:= {\mathcal {P}}(X)$

Based on which, is it reasonable to claim the power set of the natural numbers, ${\mathcal {P}}(\mathbb{N})$, is the largest possible σ-algebra on $\mathbb{N}$?

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    $\begingroup$ As your quote says: The power set is the largest $\sigma$-algebra for every set. Since $\mathbb{N}$ is a set, your claim is true. $\endgroup$ – John Aug 6 at 7:02
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    $\begingroup$ Yes.First of all, $\mathcal{P}(\mathbb N)$ is a $\sigma-$algebra. If $\Sigma$ is any $\sigma-$algebra on $\mathbb N$, then we have $\Sigma\subset\mathcal{P}(\mathbb N)$ by definition. So $\mathcal{P}(\mathbb N)$ is the largest $\sigma-$algebra on $\mathbb N$. $\endgroup$ – Feng Shao Aug 6 at 7:09

It is reasonable, provided you use a reasonable definition of biggest. $\mathcal{P}(\mathbb{N})$ is the biggest in the sense that for any $\sigma$-algebra $\Sigma$ over $\mathbb{N}$ there is a natural inclusion map $i: \Sigma \rightarrow \mathcal{P}(\mathbb{N})$. Don't try to define 'biggest' via size or cardinality, that could get messy.


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