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

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 $$\mathbb{N}$$?

• 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. – John Aug 6 at 7:02
• 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 $\mathbb N$. – 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.