Countable Subset of Minimal Uncountable Well-Ordered Set Has Upper Bound
Theorem
Let $\Omega$ denote the minimal uncountable well-ordered set.
Let $\omega$ be a countable subset of $\Omega$.
Then $\omega$ has an upper bound in $\Omega$.
Proof
By the minimal uncountable well-ordered set, $\Omega$ is uncountable
Then $\omega \ne \Omega$, for the former is countable and the latter is not.
Consider the union:
- $\ds \bigcup_{x \mathop \in \omega} S_x$
of initial segments $S_x$ in $\omega$.
By the definition of $\Omega$, any $S_x$ is countable.
Thus $\ds \bigcup_{x \mathop \in \omega} S_x$ 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:
- $\ds \bigcup_{x \mathop \in \omega} S_x = S_y$
Such a $y$ is an upper bound for $\omega$ by the definition of an initial segment.
$\blacksquare$
Axiom of Choice
This theorem depends on the Axiom of Choice, by way of Countable Union of Countable Sets is Countable.
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.
Sources
- 1984: Gerald B. Folland: Real Analysis: Modern Techniques and their Applications $\S P.19$