Ordinal is Less than Successor
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Theorem
Let $x$ be an ordinal.
Let $x^+$ denote the successor of $x$.
Then:
- $x \in x^+$
- $x \subset x^+$
Proof
$x$ is an ordinal and so by definition is also a set.
- $x \in \left({x \cup \left\{ {x}\right\} }\right) \land x \subset \left({x \cup \left\{ {x}\right\} }\right)$
so by applying the definition of a successor set:
- $x \in x^+ \land x \subset x^+$
$\blacksquare$
Sources
- 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 7.23$