Ordinal is Proper Subset of Successor
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Theorem
Let $\alpha$ be an ordinal.
Then:
- $\alpha \subsetneqq \alpha^+$
where $\alpha^+$ denotes the successor set of $\alpha$.
That is:
- $\alpha$ is a proper subset of $\alpha^+$.
Proof
Aiming for a contradiction, suppose $\alpha = \alpha^+$.
By definition:
- $\alpha^+ = \alpha \cup \set \alpha$
and so:
- $\alpha \subseteq \alpha^+$
and:
- $\alpha \in \alpha^+$
which leads to:
- $\alpha \in \alpha$
But from Ordinal is not Element of Itself:
- $\alpha \notin \alpha$
Hence by Proof by Contradiction:
- $\alpha \ne \alpha^+$
But we have:
- $\alpha \subseteq \alpha^+$
Hence:
- $\alpha \subsetneqq \alpha^+$
$\blacksquare$
Sources
- 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $5$: Ordinal Numbers: $\S 1$ Ordinal numbers: Corollary $1.9$