Existence of Minimal Uncountable Well-Ordered Set/Proof Using Choice

Proof
By the axiom of powers, there exists the power set $\mathcal P \left({\N}\right)$, where $\N$ is the set of natural numbers.

By Power Set of Natural Numbers Not Countable, this set is uncountable

By the well-ordering theorem, $\mathcal P \left({\N}\right)$ can be endowed with a well-ordering.

Denote such an ordering with the symbol $\preccurlyeq$.

Let $\mathcal P \left({\N}\right)_a$ denote the initial segments of $\mathcal P \left({\N}\right)$ determined by $a \in \mathcal P \left({\N}\right)$

Suppose $\left({P \left({\N}\right),\preccurlyeq}\right)$ has the property:


 * $\mathcal P \left({\N}\right)_a$ is countable for every $a \in \mathcal P \left({\N}\right)$

Then set $\Omega = \mathcal P\left({\N}\right)$.

Otherwise, suppose $\left({P \left({\N}\right),\preccurlyeq}\right)$ does not have the above property.

Consider the subset of $\mathcal P\left({\N}\right)$


 * $P \subseteq \left\{ { a \in \mathcal P\left({\N}\right) : \mathcal P \left({\N}\right)_a \text{ is uncountable} } \right\}$

Then $P$ has a smallest element, by the definition of a well-ordered set. Call such an element $a_0$.

That is, $a_0 \in \mathcal P \left({\N}\right)$ is the smallest $a$ such that $\mathcal P \left({\N}\right)_{a_0}$ is uncountable.

Then the segment $\mathcal P \left({\N}\right)_{a_0}$ is itself uncountable, by virtue of $a_0$ being in $P$.

Thus every initial segment in $\mathcal P \left({\N}\right)_{a_0}$ is countable, because it is not uncountable.

Then set $\Omega = \mathcal P \left({\N}\right)_{a_0}$