Countable Subset of Minimal Uncountable Well-Ordered Set Has Upper Bound

Theorem
Let $\Omega$ denote the set of countable ordinals.

Let $\omega$ be a countable subset of $\Omega$.

Then $\omega$ has an upper bound in $\Omega$.

Proof
By the definition of the set of countable ordinals, $\Omega$ is uncountable

Then $\omega \ne \Omega$, for the former is countable and the latter is not.

Consider the union:


 * $\displaystyle \bigcup_{x \mathop \in \omega} S_x$

of initial segments $S_x$ in $\omega$.

By the definition of $\Omega$, any $S_x$ is countable.

Thus $\displaystyle \bigcup_{x \mathop \in \omega} S_x$ is a countable union of countable sets.

This union is also countable, and so also cannot be all of $\Omega$, as $\Omega$ is uncountable.

From Union of Initial Segments is Initial Segment or All of Woset, there exists some $y \in \Omega$ such that:


 * $\displaystyle \bigcup_{x \mathop \in \omega} S_x = S_y$

Such a $y$ is an upper bound for $\omega$ by the definition of an initial segment.