Finite Ordinal is not Subset of one of its Elements

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $n$ be a finite ordinal.


Then:

$\nexists x \in n: n \subseteq x$

that is, $n$ is not a subset of one of its elements.


Proof

Let $S$ be the set of all those finite ordinals $n$ which are not a subset of any of its elements.

That is:

$n \in S \iff n \in \omega \land \forall x \in n: n \nsubseteq x$

We know that $0 = \varnothing$ is not a subset of any of its elements, as $\varnothing$ by definition has no elements.

So $0 \in S$.


Now suppose $n \in S$.

From Set is Subset of Itself:

$n \subseteq n$

But as $n \in S$ it follows by definition of $S$ that:

$n \notin n$

By definition of the successor of $n$, it follows that:

$n^+ \nsubseteq n$

Now from Subset Relation is Transitive:

$n^+ \subseteq x \implies n \subseteq x$

But since $n \in S$ it follows that:

$x \notin n$

So:

$n^+ \nsubseteq n$

and:

$\forall x \in n: n^+ \nsubseteq x$

So $n^+$ is not a subset of any of its elements.

That is:

$n^+ \in S$

So by the Principle of Mathematical Induction:

$S = \omega$

Hence the result.

$\blacksquare$


Sources