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