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:


 * $\neg \exists m \in \N: \paren {n < m < n^+}$

That is:
 * If $m \le n \le m^+$, then $m = n$ or $m = n^+$.

Proof using Minimal Infinite Successor Set
such an ordinal $y$ exists.

Then, by Ordering on Ordinal is Subset Relation:


 * $x \in y$

and from Transitive Set is Proper Subset of Ordinal iff Element of Ordinal:


 * $y \in x^+$

Applying the definition of a successor set, we have:


 * $y \in x \lor y = x$

But this creates a membership loop, in that:


 * $x \in y \in x \lor x \in x$

By No Membership Loops, we have created a contradiction.

The result follows from Proof by Contradiction.