Ordinal is not Element of Itself/Proof 3

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Theorem

Let $x$ be an ordinal.

Then:

$x \notin x$


Proof

Let $\On$ denote the class of all ordinals.

From Class of All Ordinals is Minimally Superinductive over Successor Mapping, $\On$ is superinductive.

Hence we can use the Principle of Superinduction.


By Zero is Smallest Ordinal, $0$ is the smallest element of $\On$.

We identify the natural number $0$ via the von Neumann construction of the natural numbers as:

$0 := \O$

From Empty Set is Ordinary:

$\O \notin \O$

and so:

$0 \notin 0$

$\Box$


Let $x$ be an ordinal for which it has been established that:

$x \notin x$

Then by Successor Set of Ordinary Transitive Set is Ordinary‎:

$x^+ \notin x^+$

$\Box$


It remains to be shown that if $C$ is a chain of ordinals all of which are ordinary, then $\bigcup C$ is ordinary.


Let $C$ be a chain of ordinals.

From Union of Set of Ordinals is Ordinal, $\bigcup C$ is itself an ordinal.


Let $\alpha = \bigcup C$.

Aiming for a contradiction, suppose $\bigcup C$ is not ordinary.

That is, suppose that $\alpha \in \alpha$.

Then:

$\alpha \in \bigcup C$

Hence for some $\beta \in C$:

$\alpha \in \beta$

But:

$\beta \in C$

and so:

$\beta \subseteq \bigcup C$

That is:

$\beta \subseteq \alpha$

Because $\alpha \in \beta$ and Ordinal is Transitive:

$\alpha \subseteq \beta$

Hence by definition of set equality:

$\alpha = \beta$

That is:

$\beta \in \beta$

Thus some element of $C$, that is, $\beta$, is an element of itself.

That is, not all elements of $C$ are ordinary.

This contradicts our assertion that all elements of $C$ are ordinary.

Hence by Proof by Contradiction we have that if $C$ is a chain of ordinals all of which are ordinary, then $\bigcup C$ is ordinary:

$\bigcup C \notin \bigcup C$

$\Box$


The result then follows by the Principle of Superinduction.

$\blacksquare$


Sources

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