Can we say N is strictly a subset of P(N)?

[AATroop]

Essentially, do we know \mathbb{N}\subset{P}\mathbb{(N)}?

 

[micromass]

A lot depends on your definition of \mathbb{N}. But if we take the usual set theoretic definition, then the inclusion is indeed true.

What is the set theoretic definition of \mathbb{N}? It goes as follows:

• Define 0=\emptyset
• If 0,...,n are defined, then define n+1=\{0,...,n\}
• Define \mathbb{N}=\{0,1,2,...\}

Now we can check that \mathbb{N}\subseteq \mathcal{P}(\mathbb{N}). First of all, \emptyset is a subset of \mathbb{N} and thus \emptyset\in \mathcal{P}(\mathbb{N}). Furthermore, \{0,...,n\} is a subset of \mathbb{N}, and thus n+1=\{0,...,n\}\in \mathcal{P}(\mathbb{N}).
This shows that \mathbb{N}\subseteq \mathcal{P}(\mathbb{N}). The inclusion is strict since \{2\} is a subset of \mathbb{N}, but not an element.