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

Proof
By the axiom of powers, there exists the power set $\powerset \N$, where $\N$ is the set of natural numbers.

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

By the well-ordering theorem, $\powerset \N$ can be endowed with a well-ordering.

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

Let $\powerset \N_a$ denote the initial segments of $\powerset \N$ determined by $a \in \powerset \N$

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


 * $\powerset \N_a$ is countable for every $a \in \powerset \N$

Then set $\Omega = \powerset \N$.

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

Consider the subset of $\powerset \N$:


 * $P \subseteq \set {a \in \powerset \N : \powerset \N_a \text{ is uncountable} }$

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

That is, $a_0 \in \powerset \N$ is the smallest $a$ such that $\powerset \N_{a_0}$ is uncountable.

Then the segment $\powerset \N_{a_0}$ is itself uncountable, by virtue of $a_0$ being in $P$.

Thus every initial segment in $\powerset \N_{a_0}$ is countable, because it is not uncountable.

Then set $\Omega = \powerset \N_{a_0}$