Cofinal to Zero iff Ordinal is Zero

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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.

$\blacksquare$


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