Nonlimit Ordinal Cofinal to One
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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:
- $\operatorname{cof} \left({x, 1}\right)$
$\blacksquare$
Sources
- 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 10.54$