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

1. Oct 26, 2012

### AATroop

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

2. Oct 26, 2012

### micromass

Staff Emeritus
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.