No Natural Number between Number and Successor
Theorem
Let $\N$ be the natural numbers.
Let $n \in \N$.
Then no natural number $m$ exists strictly between $n$ and its successor $n + 1$:
- $\neg \exists m \in \N: \paren {n < m < n + 1}$
That is:
- If $n \le m \le n + 1$, then $m = n$ or $m = n + 1$.
Proof using Minimally Inductive Set
We use the model that $\N \cong \omega$ where $\omega$ is the minimally inductive set.
Aiming for a contradiction, suppose such an ordinal $m$ exists.
Then, by Ordering on Ordinal is Subset Relation:
- $n \in m$
and from Transitive Set is Proper Subset of Ordinal iff Element of Ordinal:
- $m \in n^+$
Applying the definition of a successor set, we have:
- $n \in m \lor n = m$
But this creates a membership loop, in that:
- $m \in n \in m \lor m \in m$
By No Membership Loops, we have created a contradiction.
The result follows from Proof by Contradiction.
$\blacksquare$
Proof using Von Neumann Construction
Let $\N$ be defined as the von Neumann construction $\omega$.
By definition of the ordering on von Neumann construction:
- $m \le n \iff m \subseteq n$
From Von Neumann Construction of Natural Numbers is Minimally Inductive, $\omega$ is minimally inductive class under the successor mapping.
The result follows from the Sandwich Principle:
- $\forall m, n \in \omega: m \subseteq n \subseteq m^+ \implies m = n \lor n = m^+$
where $m^+$ is identified as $m + 1$.
$\blacksquare$