# Given $2$ ordinals $\alpha$ and $\beta$, how to contruct an ordinal contains them as elements?

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 $α ≤ β ⇔ α ∈ β \lor α = β.$

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.
• Sorry but so far I cannot see why $\alpha\subseteq \gamma\Rightarrow \alpha\in \gamma$, which is a direction of "equivalent". – Y.X. Mar 22 '17 at 5:19
• That is, $\alpha, \gamma$ ordinals satisfy $\alpha \subseteq \gamma \implies (\alpha \in \gamma \lor \alpha = \gamma)$. – hardmath Mar 22 '17 at 19:12