Initial Segment of Ordinal is Ordinal

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $S$ be an ordinal, and suppose that $a \in S$.


Then the initial segment $S_a = a$ of $S$ determined by $a$ is also an ordinal.


In other words, every element of an ordinal is also an ordinal.


Proof

By Subset of Well-Ordered Set is Well-Ordered, $S_a$ is well-ordered.


Suppose that $b \in S_a$.

From Ordering on Ordinal is Subset Relation, and the definition of an initial segment, it follows that $b \subset a$.


Then:

\(\displaystyle \paren {S_a}_b\) \(=\) \(\displaystyle \set {x \in S_a: x \subset b}\) Definition of Initial Segment
\(\displaystyle \) \(=\) \(\displaystyle \set {x \in S: x \subset a \land x \subset b}\) Definition of Initial Segment
\(\displaystyle \) \(=\) \(\displaystyle \set {x \in S: x \subset b}\) as $b \subset a$
\(\displaystyle \) \(=\) \(\displaystyle S_b\) Definition of Initial Segment
\(\displaystyle \) \(=\) \(\displaystyle b\) as $S$ is an ordinal


The result follows from the definition of an ordinal.

$\blacksquare$


Sources