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

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.

Proof

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

By Zermelo's 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 $\struct {\powerset \N, \preccurlyeq}$ has the property:

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

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

Otherwise, suppose $\struct {\powerset \N, \preccurlyeq}$ 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 initial 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:

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

$\blacksquare$

Axiom of Choice

This theorem depends on the Axiom of Choice, by way of Zermelo's Well-Ordering Theorem.

Because of some of its bewilderingly paradoxical implications, the Axiom of Choice is considered in some mathematical circles to be controversial.

Most mathematicians are convinced of its truth and insist that it should nowadays be generally accepted.

However, others consider its implications so counter-intuitive and nonsensical that they adopt the philosophical position that it cannot be true.