# No Infinitely Descending Membership Chains

## Theorem

Let $\omega$ denote the minimally inductive set.

Let $F$ be a mapping whose domain is $\omega$.

Then:

$\exists n \in \omega: \map F {n^+} \notin \map F n$

### Corollary

There cannot exist a sequence $\sequence {x_n}$ whose domain is $\N_{\gt 0}$ such that:

$\forall n \in \N_{>0}: x_{n + 1} \in x_n$

## Axiom of Foundation

This theorem depends on the Axiom of Foundation.

Most mathematicians accept the Axiom of Foundation, but theories that reject it, or negate it, have found applications in Computer Science and Linguistics.

## Proof

Let $F$ be a mapping whose domain is $\omega$.

By the axiom of replacement, the image of $F$ is a set.

Let the image of $F$ be denoted $\map \WW F$.

Then:

 $\ds$  $\ds \exists x \in \map \WW F: \paren {\map \WW F \cap x} = \O$ Axiom of Foundation $\ds$ $\leadsto$ $\ds \exists x: \paren {\paren {\map \WW F \cap x} = \O \land \exists n \in \omega: x = \map F n}$ as $x \in \map \WW F$, $x = \map F n$ for some $n \in \omega$ $\ds$ $\leadsto$ $\ds \exists n \in \omega: \paren {\map \WW F \cap \map F n} = \O$ Logical manipulation eliminating $x$

But:

$\map F {n^+} \in \map \WW F$

So:

$\map F {n^+} \notin \map F n$

$\blacksquare$