# Continuum equals Cardinality of Power Set of Naturals It has been suggested that this page or section be merged into Cardinality of Power Set of Natural Numbers Equals Cardinality of Real Numbers. (Discuss)

## Theorem

$\mathfrak c = \card {\powerset N}$

where

$\powerset N$ denotes the power set of $\N$
$\card {\powerset N}$ denotes the cardinality of $\powerset N$
$\mathfrak c = \card \R$ denotes the continuum.

## Proof

By Reals are Isomorphic to Dedekind Cuts there exists bijection:

$f: \R \to \mathscr D$

where:

$\mathscr D$ denotes the set of all Dedekind cuts of $\struct {\Q, \le}$.

Dedekind's cuts are subsets of $\Q$.

Therefore by definition of power set:

$\mathscr D \subseteq \powerset \Q$
$\card {\mathscr D} \le \card {\powerset \Q}$
$\Q$ is countably infinite.

Then by definition of countably infinite there exists a bijection:

$g: \Q \to \N$

By definition of set equivalence:

$\Q \sim \N$

Hence by definition of cardinality:

$\card \Q = \card \N$
$\card {\powerset \Q} = \card {\powerset \N}$

By definition of set equivalence:

$\R \sim \mathscr D$

Hence by definition of cardinality:

$\card \R = \card {\mathscr D}$

Thus:

$\mathfrak c \le \card {\powerset \N}$

Define a mapping $h: \map {\operatorname {Fin} } \N \times \powerset \N \to \R^+$:

$\forall F \in \map {\operatorname {Fin} } \N, A \in \powerset \N: \map h {F, A} = \displaystyle \sum_{i \mathop \in F} 2^i + \sum_{i \mathop \in A} \paren {\frac 1 2}^i$

where $\map {\operatorname {Fin} } \N$ denotes the set of all finite subsets of $\N$.

A pair $\tuple {F, A}$ corresponds to binary denotation of a real number $\map h {F, A}$.

It means that $h$ is a surjection.

$\card {\map {\operatorname {Fin} } \N \times \powerset \N} \le \card {\R^+}$

By definition of subset:

$\map {\operatorname {Fin} } \N \subseteq \powerset \N$
$\card {\map {\operatorname {Fin} } \N} \le \card {\powerset \N}$
 $\displaystyle \card {\map {\operatorname {Fin} } \N \times \powerset \N}$ $=$ $\displaystyle \max \set {\card {\map {\operatorname {Fin} } \N}, \card {\powerset \N} }$ Cardinal Product Equal to Maximum $\displaystyle$ $=$ $\displaystyle \card {\powerset \N}$

Because $\R^+ \subseteq \R$, we have by Subset implies Cardinal Inequality:

$\card {\R^+} \le \card \R$

Thus:

$\card {\powerset \N} \le \mathfrak c$

Hence the result:

$\mathfrak c = \card {\powerset \N}$

$\blacksquare$