TheoremBase

Elements of a Segment Are Natural Numbers, and Mixed Transitivity of the Order on the Natural Numbers

Every element of the natural numbers with zero lying between 1 and a set n, in particular every element of a segment [n], is a natural number; and on the natural numbers a strict and a non-strict inequality in a chain give a strict inequality.

Statement

In the setting of The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion, let ≤\le and << be the order and the strict order of N0\mathbb{N}_{0}, as in The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion (v), and let [n][n] be as in Intervals of Natural Numbers (ii).

Let nn be a set, not necessarily an element of N0\mathbb{N}_{0}. Every k∈N0k\in\mathbb{N}_{0} with 1≤k≤n1\le k\le n is a natural number, k∈Nk\in\mathbb{N}; only the lower bound 1≤k1\le k is needed for this, the upper bound k≤nk\le n plays no role. In particular, for n∈N0n\in\mathbb{N}_{0}, every element of [n][n] is a natural number.

Let k,l,m∈Nk,l,m\in\mathbb{N}. If k<lk<l and l≤ml\le m, then k<mk<m; and if k≤lk\le l and l<ml<m, then k<mk<m.

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