Existence of Minimal Uncountable Well-Ordered Set

Theorem
There exists a minimal uncountable well-ordered set.

That is, there exists an uncountable well-ordered set $\Omega$ with the property that every initial segment in $\Omega$ is countable.

Corollary
The cardinality of $\Omega$ satisfies:


 * $\operatorname{card} \left({\N}\right) < \operatorname{card} \left({\Omega}\right) \le \mathfrak c$

where $\mathfrak c$ is the cardinality of the continuum.

Proof using choice
By the axiom of powers, there exists the power set $\mathcal P \left({\N}\right)$.

This set is uncountable.

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

Let $\left({P \left({\N}\right),\preccurlyeq}\right)$ be some well-ordering on $P \left({\N}\right)$.

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

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


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

then we have found the sought-after minimal uncountable well-ordered set.

Otherwise, by the definition of a well-ordering, there is a smallest element $a_0 \in \mathcal P \left({\N}\right)$ such that $\mathcal P \left({\N}\right)_{a_0}$ is uncountable.

Then the segment $\mathcal P \left({\N}\right)_{a_0}$ is itself uncountable, but every initial segment in $\mathcal P \left({\N}\right)_{a_0}$ is countable.

Set $\Omega = \mathcal P \left({\N}\right)_{a_0}$.

Proof without using choice
By the axiom of powers, there exists the power set $\mathcal P \left({\N}\right)$.

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

We construct a well-ordering $\left({P \left({\N}\right), \preccurlyeq}\right)$ that has the desired defining properties of $\Omega$.

Proof of Corollary
That $\operatorname{card} \left({\N}\right) < \operatorname{card} \left({\Omega}\right)$ follows from $\left({\Omega}\right)$ is uncountable by definition of minimal uncountable well-ordered set.

$\operatorname{card} \left({\Omega}\right) \le \operatorname{card}\left({\mathcal P \left({\N}\right)}\right)$ follows the construction of the proof, as the $\Omega$ constructed is a subset of $\mathcal P \left({\N}\right)$.

That $\operatorname{card} \left({\mathcal P \left({\N}\right) }\right) = \mathfrak c$ is showed in Cardinality of Power Set of Natural Numbers Equals Cardinality of Real Numbers.

Also see

 * Minimal Uncountable Well-Ordered Set Unique up to Isomorphism