# 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:

$\operatorname{cof} \left({ x,0 }\right)$
$\operatorname{cof} \left({ 0,x }\right)$
$x = 0$

## Proof

### $\operatorname{cof} \left({ x,0 }\right) \implies x = 0$

If $\operatorname{cof} \left({ x,0 }\right)$, then there is a function $f : x \to 0$.

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

But then, $f\left({a}\right) \in 0$, which contradicts the definition of the empty set.

Therefore, $x = 0$.

$\Box$

### $x = 0 \implies \operatorname{cof} \left({ 0,x }\right)$

Follows directly from Cofinal Ordinal Relation is Reflexive.

$\Box$

### $\operatorname{cof} \left({ 0,x }\right) \implies \operatorname{cof} \left({ x,0 }\right)$

If $\operatorname{cof} \left({ 0,x }\right)$, then $x \le 0$ by the definition of cofinal.

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

$\operatorname{cof} \left({ x,0 }\right)$ follows directly from Cofinal Ordinal Relation is Reflexive.

$\blacksquare$