Cofinal to Zero iff Ordinal is Zero

Theorem

Let $x$ be an ordinal.

Let $\operatorname{cof}$ denote the cofinal relation.

Let $0$ denote the zero ordinal.

Then the following are equivalent:

$\map {\operatorname{cof} } {x, 0}$

$\map {\operatorname{cof} } {0, x}$

$x = 0$

Proof

$\map {\operatorname{cof} } {x, 0} \implies x = 0$

If $\map {\operatorname{cof} } {x, 0}$, then there is a function $f : x \to 0$.

If $x \ne 0$, then $x$ has an element $a$.

But then, $\map f a \in 0$, which contradicts the definition of the empty set.

Therefore, $x = 0$.

$\Box$

$x = 0 \implies \map {\operatorname{cof} } {0, x}$

Follows directly from Cofinal Ordinal Relation is Reflexive.

$\Box$

$\map {\operatorname{cof} } {0, x} \implies \map {\operatorname{cof} } {x, 0}$

If $\map {\operatorname{cof} } {0, x}$, then $x \le 0$ by the definition of cofinal.

By Subset of Empty Set, $x = 0$.

$\map {\operatorname{cof} } {x, 0}$ follows directly from Cofinal Ordinal Relation is Reflexive.

$\blacksquare$

Sources

• 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 10.53 \ (1)$