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Suppose $\alpha, \beta$ are ordinals, as ordinals are sets of ordinals, it seems that it is possible to construct an ordinal that contains $\alpha$ and $\beta$ as elements.

I am not able to construct such an ordinal so far. If it is possible, could someone give an explicit construction with justification? Thanks so much!

definition of ordinal:

transitive: $x ⊆ S$ for every $x ∈ S$.

A total order $≤$ on a set S is said to be a well-ordering if for every non-empty subset $A ⊆ S$, there exists an element $m ∈ A$ such that $ ∀x ∈ A,m\le x$

An ordinal number is a set that is transitive and is well-ordered by the relation $α ≤ β ⇔ α ∈ β \lor α = β. $

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Ordinals are transitive sets, so $\alpha \in \gamma$ is equivalent to $\alpha \subsetneq \gamma$. So take their union, and show it's an ordinal. Then take the successor.

That a union of ordinals (in your definition) is an ordinal can be found here. So $\gamma:=\alpha \cup \beta$ is an ordinal, such that $\alpha \subseteq \gamma$ and $\beta \subseteq \gamma$. To ensure that $\alpha \in \delta$ we need $\alpha \subsetneq \delta$, so define $\delta = \gamma + 1 =\gamma \cup \{\gamma\}$ and this contains $\alpha$ and $\beta$ as elements.

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  • $\begingroup$ Sorry but so far I cannot see why $\alpha\subseteq \gamma\Rightarrow \alpha\in \gamma$, which is a direction of "equivalent". $\endgroup$ – Y.X. Mar 22 '17 at 5:19
  • $\begingroup$ @Y.X. The proof will depend on the exact definition of ordinal you're using $\endgroup$ – Henno Brandsma Mar 22 '17 at 5:21
  • $\begingroup$ @Could you please look at the edited question where I give a definition? $\endgroup$ – Y.X. Mar 22 '17 at 5:29
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    $\begingroup$ @Y.X. Successor of union is still my answer $\endgroup$ – Henno Brandsma Mar 22 '17 at 6:00
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    $\begingroup$ That is, $\alpha, \gamma$ ordinals satisfy $\alpha \subseteq \gamma \implies (\alpha \in \gamma \lor \alpha = \gamma)$. $\endgroup$ – hardmath Mar 22 '17 at 19:12

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