Power Set of Natural Numbers has Cardinality of Continuum/Proof 2

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:
 * $\mathscr D \subseteq \powerset \Q$

By Subset implies Cardinal Inequality:
 * $\card {\mathscr D} \le \card {\powerset \Q}$

By Rational Numbers are Countably Infinite:
 * $\Q$ 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:
 * $\card \Q = \card \N$

Then by Cardinality of Power Set is Invariant:
 * $\card {\powerset \Q} = \card {\powerset \N}$

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

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.

By Surjection iff Cardinal Inequality:
 * $\card {\map {\operatorname {Fin} } \N \times \powerset \N} \le \card {\R^+}$

By definition of subset:
 * $\map {\operatorname {Fin} } \N \subseteq \powerset \N$

Then by Subset implies Cardinal Inequality:
 * $\card {\map {\operatorname {Fin} } \N} \le \card {\powerset \N}$

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