If $\alpha$ is an ordinal, proving that $\alpha\cup\{\alpha\}$ is an ordinal.

Question

I refer to pg.4 of this article.

Assuming $\alpha$ is an ordinal, we have to prove $\alpha\cup \{\alpha\}$ or $\alpha +1$ is an ordinal.

1. Isn't this obvious from the construction of ordinals? As per the construction given in the article, for any ordinal $\beta$, the next ordinal is $\beta\cup\{\beta\}$. So is any such explicit proof required, as is given in the article?

2. I quote: "$\alpha +1$ is transitive, for if $y\in\alpha +1$ then either $y=\alpha$ and $\alpha\subset \alpha +1$, or $y\in\alpha$." I don't understand how this follows from the properties of ordinals given on pg.3. I feel this is a proof of the fact that $\alpha +1$ is transitive by assuming that $\alpha +1$ is transitive. One may refer to definition 7 on pg.3

EDIT: Could someone also kindly outline the suffficient conditions for proving that a number is an ordinal? The artice suggests transitivity and strict ordering are suffcient conditions, or maybe I'm reading it wrong.

Thanks in advance!

Answer 1

You need to discern between "obvious equivalence" and definitions. The definition of an ordinal is a set which is transitive and well-ordered by $\in$.

If $\alpha$ is assumed to be an ordinal this means that it is a transitive set and well-ordered by $\in$. Now you want to prove that $\alpha+1=\alpha\cup\{\alpha\}$ also satisfies the same properties.

This is a very simple proof, yes. But it is required regardless. Even more so because this is supposed to be an paper detailing the construction of ordinals to people which are less familiar with them.

To the second question, we assume that $\alpha$ is transitive. Therefore if $x\in\alpha\cup\{\alpha\}$, either $x\in\alpha$ and therefore $x\subseteq\alpha\subseteq\alpha\cup\{\alpha\}$, or $x=\alpha$ and then trivially $x\subseteq\alpha\cup\{\alpha\}$. Therefore $\alpha+1$ is transitive whenever $\alpha$ is.

As for the equivalent conditions for being an ordinal:

1. $x$ is an ordinal if $x$ is a transitive set which is well-ordered by $\in$. Assuming the axiom of regularity it suffices to require $\in$ to be a linear order instead.
2. $x$ is an ordinal if $x$ is a transitive set, and all its members are transitive sets. This definition requires the axiom of regularity to holds as well.

Answer 2

In the linked article an ordinal is defined (Definition 8) as (1) "a transitive set" (defined just above, Definition 7) that is (2) "well-ordered by the relation $x \lt y$ if[f] $x \in y$".

So the proof that $\alpha \cup \{\alpha\}$ is an ordinal (assuming $\alpha$ is) should be done in the first instance by applying Definition 8, which is probably not "per the construction" you have in mind.

The sufficient conditions you request are given in Definition 8: transitivity and well-ordering by membership relation.