# Nonlimit Ordinal Cofinal to One

## Theorem

Let $x$ be a nonlimit non-empty ordinal.

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

Let $1$ denote the ordinal one.

Then:

$\operatorname{cof} \left({x, 1}\right)$

## Proof

Since $1 = 0^+$, $1$ is not a limit ordinal.

Let $\le$ denote the subset relation.

It follows that $0 < 1$ by Ordinal is Less than Successor.

Moreover, $1 \le x$ follows by the fact that $0 < x$ and Successor of Element of Ordinal is Subset.

Thus we have $0 < 1 \le x$ and so by Condition for Cofinal Nonlimit Ordinals:

$\operatorname{cof} \left({x, 1}\right)$

$\blacksquare$