Ordinal is Subset of Successor
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Theorem
Let $x$ and $y$ be ordinals.
Let $x^+$ denote the successor of $x$.
Then:
- $x \subseteq y^+ \iff \left({x \subseteq y \lor x = y^+}\right)$
Proof
Let $A \subset B$ denote that $A$ is a proper subset of $B$.
Let $A \in B$ denote that $A$ is an element of $B$.
From Transitive Set is Proper Subset of Ordinal iff Element of Ordinal, $\subset$ and $\in$ can be used interchangeably.
Thus:
| \(\ds x \subseteq y\) | \(\leadsto\) | \(\ds x \subseteq y^+\) | Ordinal is Less than Successor | |||||||||||
| \(\ds x = y^+\) | \(\leadsto\) | \(\ds x \subseteq y^+\) | Definition 2 of Set Equality |
| \(\ds x \subseteq y^+\) | \(\leadsto\) | \(\ds \paren {x \subset y^+ \lor x = y^+}\) | ||||||||||||
| \(\ds x \subset y^+\) | \(\leadsto\) | \(\ds x \in y^+\) | Transitive Set is Proper Subset of Ordinal iff Element of Ordinal | |||||||||||
| \(\ds \) | \(\leadsto\) | \(\ds \paren {x = y \lor x \in y}\) | Definition of Successor Set | |||||||||||
| \(\ds \) | \(\leadsto\) | \(\ds x \subseteq y\) | Transitive Set is Proper Subset of Ordinal iff Element of Ordinal |
$\blacksquare$