Successor of Ordinal Smaller than Limit Ordinal is also Smaller/Proof 1

Theorem

Let $\On$ denote the class of all ordinals.

Let $\lambda \in \On$ be a limit ordinal.

Then:

$\forall \alpha \in \On: \alpha < \lambda \implies \alpha^+ < \lambda$

Proof

Let $\lambda$ be a limit ordinal such that $\alpha < \lambda$.

From Successor of Element of Ordinal is Subset

Then as $\alpha^+$ is the successor set of $\alpha$ it follows that:

$\alpha^+ \le \lambda$

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But as $\lambda$ is not a successor ordinal:

$\alpha^+ \ne \lambda$

Hence:

$\alpha^+ < \lambda$

$\blacksquare$

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: Theorem $1.19$