What is bigger, $p(\mathbb{N})$ or $\mathbb{R}$?

Question

I know that $|p(\mathbb{N})|>|\mathbb{N}|$, and that $|\mathbb{R}|>|\mathbb{N}|$, and I wonder whether $|p(\mathbb{N})|>|\mathbb{R}|$ or not.

What I tried so far: I found the function from $\mathbb{R}$ to $p(\mathbb{Q})$ defined by $f(x)=\{q\in \mathbb{Q}|q<x\}$, which I am quite sure to be injective function, but not onto. As $|\mathbb{Q}|=|\mathbb{N}|$, also $|p(\mathbb{Q})|=|p(\mathbb{N})|$, so I inferred that $|p(\mathbb{N})|\ge |\mathbb{R}|$. But are they equal?

In this, $p(A)$ is the power set of A, denoted also by $2^A$ and defined as is the set of all subsets of A.

This seems related: http://math.stackexchange.com/questions/209396/is-2-mathbbn-mathbbr — Martin Sleziak, Feb 18 '14 at 07:48
Another related question: http://math.stackexchange.com/questions/658112/why-does-mathbbr-have-the-same-cardinality-as-mathcalp-mathbbn — Martin Sleziak, Feb 18 '14 at 07:53
@Martin: Don't be afraid calling the child by its name. Duplicates, not "related". — Asaf Karagila, Feb 18 '14 at 07:57

score 14 · Accepted Answer · answered Feb 17 '14 at 11:54

14

You have an injection $\mathbb{R}\hookrightarrow p(\mathbb{N})$. To see that $\lvert \mathbb{R}\rvert = \lvert p(\mathbb{N})\rvert$, you now need an injection $p(\mathbb{N}) \hookrightarrow \mathbb{R}$.

For $M\subset \mathbb{N}$, consider for example

$$v(M) = \sum_{n\in M} 3^{-n}.$$

answered Feb 17 '14 at 11:54

Daniel Fischer

206,697

For which numbers other than $3$ does it work? Does it work for all? – Anna Feb 17 '14 at 15:34
1

It works for all $b > 2$. For $b = 2$, we get a surjection $p(\mathbb{N}) \to [0,1]$ (or $[0,2]$, depending on whether you have $0\in \mathbb{N}$) which is not injective, since $2^{-k} = \sum_{k+1}^\infty 2^{-n}$. For $b > 2$, we have $b^{-k} > \sum_{k+1}^\infty b^{-n}$ and thus injectivity; $b$ need not even be an integer. – Daniel Fischer Feb 17 '14 at 15:41

George Tomlinson · Answer 2 · 2014-02-18T11:08:12.593

Alternatively observe that the existence of the following injection from $p(\mathbb{Q})$ to $\mathbb{R}$ together with the fact that there exists an injection from $\mathbb{R}$ to $p(\mathbb{Q})$ (i.e. the one you give) shows that $\lvert \mathbb{R}\rvert = \lvert p(\mathbb{Q})\rvert$:

For $M\subset \mathbb{Q}$,

$$v(M) = \sum_{i=1}^n 10^{i-1+\sum_{i=1}^n p_i } + w(M) + \sum_{i=1}^n 10^{-(i+1+ \sum_{i=1}^n q_i )}$$

where $w(M)=0.1$ if $0 \in M$, $0.2$ otherwise,

and $M = ${$\frac{p_1}{q_1}, \frac{p_2}{q_2},..., \frac{p_n}{q_n}$}.

For example, $v(${$\frac{1}{3}, \frac{2}{5}$}$)=10100.20001000001$

What is bigger, $p(\mathbb{N})$ or $\mathbb{R}$?

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