# No Natural Number between Number and Successor/Proof using Minimally Inductive Set

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## 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

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.

$\blacksquare$

## Sources

- 1971: Gaisi Takeuti and Wilson M. Zaring:
*Introduction to Axiomatic Set Theory*: $\S 7.25$