Initial Segment of One-Based Natural Numbers determined by Zero is Empty
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Theorem
Let $\N^*_k$ denote the initial segment of the one-based natural numbers determined by $k$:
- $\N^*_k = \set {1, 2, 3, \ldots, k - 1, k}$
Then $\N^*_0 = \O$.
Proof
From the definition of $\N^*_0$:
- $\N^*_0 = \set {n \in \N_{>0}: n \le 0}$
From the definition of one, the minimal element of $\N_{>0}$ is $1$.
From Zero Strictly Precedes One we know that $0 < 1$.
So there is no element $n$ of $\N_{>0}$ such that $n \le 0$.
Thus $\N^*_0 = \O$.
$\blacksquare$