# Proving $\mathcal P(\mathbb N) \sim (\mathcal P(\mathbb N)\setminus \{\emptyset\})$

Prove: $\mathcal P(\mathbb N) \sim (\mathcal P(\mathbb N)\setminus \{\emptyset\})$

My attempt:

One side is obvious since $\mathcal P(\mathbb N) \supset (\mathcal P(\mathbb N)\setminus \{\emptyset\})$ so $|\mathcal P(\mathbb N)| > |(\mathcal P(\mathbb N)\setminus \{\emptyset\})|$ so we know there's an injective function from the RHS to LHS.

For the other side, define $f$ to be:

$$f(X)=\begin{cases} \{1\} &,X=\emptyset \\ \{n+1\} &, X=\{n\} \\ X &, else \end{cases}$$

We just performed a linear operation on all sets $X=\{n\}$ otherwise the function is the identity, therefore it's injective.

From CSB we can conclude that $\mathcal P(\mathbb N) \sim (\mathcal P(\mathbb N)\setminus \{\emptyset\})$

Is that alright?

• It is OK, but why refer to CSB when you can exhibit a bijection? – André Nicolas Feb 4 '15 at 13:11
• @AndréNicolas haven't thought about that, is the explanation for $f$ being onto the same? – shinzou Feb 4 '15 at 13:13
• Yes, the mapping is a bijection, the old Hilbert infinite hotel trick. – André Nicolas Feb 4 '15 at 13:16

Your solution is alright. You can do it also like this:

$$\newcommand{\PP}{\mathcal{P}} \Big|\PP(\mathbb{N})\Big| = \Big|\PP(\mathbb{N}\setminus\{0\})\Big| = \Bigg|\Big\{A \cup \{0\}: A \in \PP\big(\mathbb{N}\setminus\{0\}\big)\Big\}\Bigg| \leq \Big|\PP(\mathbb{N})\setminus\{\varnothing\}\Big| \leq \Big|\PP(\mathbb{N})\Big|.$$

I hope this helps $\ddot\smile$