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

Question

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. $

Answer 1

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.