Cardinals are Totally Ordered
Corollary to Zermelo's Theorem
Every set of cardinals is totally ordered under $\le$.
Proof
Let $S$ be a set of cardinals.
From Zermelo's theorem, $S$ is well-ordered.
Let $a, b \in S$.
Consider the subset $X = \set {a, b}$ of $S$.
Since $S$ is well-ordered, $\map \inf X$ exists and belongs to $X$.
So either $\map \inf X = a$ or $\map \inf X = b$.
By definition of infimum, we have that $\map \inf X \le a$ and $\map \inf X \le b$.
It follows that either $a \le b$ or $b \le a$.
So, by definition, $S$ is totally ordered under $\le$.
$\blacksquare$
Axiom of Choice
This theorem depends on the Axiom of Choice, by way of Zermelo's Theorem (Set Theory).
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
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 8$: Theorem $8.3$: Corollary