Natural Numbers are Comparable/Strong Result
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Theorem
Let $\N$ be the natural numbers.
Let $m, n \in \N$.
Then either:
- $(1): \quad m + 1 \le n$
or:
- $(2): \quad n \le m$
Proof 1
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.
Then from Minimally Inductive Class under Progressing Mapping induces Nest, $\omega$ is a nest in which:
- $\forall m, n \in \omega: \map g x \subseteq y \lor y \subseteq x$
From the definition of $\map g x$ in this context:
- $\forall x \in \omega: \map g x = x^+$
That is:
- $\forall m, n \in \N: m + 1 \subseteq n \lor n \subseteq m$
whence the result.
$\blacksquare$
Proof 2
This theorem requires a proof. In particular: Proof using Minimally Inductive Class under Slowly Progressing Mapping is Nest by exploiting Successor Mapping is Slowly Progressing. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
Sources
- 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $3$: The Natural Numbers: $\S 5$ Applications to natural numbers: Theorem $5.2$