# Prove that elements of ordinals are ordinals.

I am aiming to prove that: if $\alpha$ is an ordinal number and $\beta\in\alpha$, then $\beta$ is an ordinal number.

The definition of ordinal is given by:

An ordinal number is a set that is transitive and is well-ordered by the relation $\alpha<\beta\Leftrightarrow\alpha\in\beta$

where transitive is defined to be:

$x\subseteq S$ for every $x\in S$

And well-ordered is defined to be:

For every non-empty subset $A\subseteq S$, there exists an element $m\in A$ such that $\forall x\in A$, $m\leq x$

Here is a solution：

Theorem 4: Elements of ordinals are ordinals.

Proof: Elements of ordinals are subsets and so are well-ordered. Let $x\in y \in z \in\alpha$ where $\alpha$ is an ordinal. Then since $\alpha$ is transitive, $x, y, z\in\alpha$. Since $\in$ is transitive on $\alpha$, $x\in z$.

I cannot understand the method of checking of transitivity. I think in order to prove the element $z\in\alpha$ is transitive, we need to pick any element $x\in z$ and by definition conclude $x\subseteq z$ rather than assume that $x\in y\in z\in \alpha$. In fact, I cannot understand the whole part of the argument about transitivity in this proof.

EDIT:NOTE: Latter I realized that the answer below has not solved my question (Sorry).This version of definition of well ordering does not mention transitivity.

EDIT:Here I am asking about how to use this version of well-orderedness to prove that $\in$ is transitive. What makes ordinal special and allows $\in$ to be transitive? Could someone please use the definition above to prove it?

• Hint: $\in$ is a well ordering in $\alpha$. – Renan Maneli Mezabarba Mar 21 '17 at 12:21
• @RenanManeliMezabarba I know that, and it is mentioned by the proof. May I please ask for an explicit solution? – PropositionX Mar 21 '17 at 12:25

We want to show that $z\in\alpha$ is transitive, i.e., if $y\in z$ then $y\subset z$. In order to prove this inclusion we take an arbitrary $x\in y$ and we show that $x\in z$. So, if $x\in y$, notice that $x,y,z\in\alpha$:

• $z\in\alpha$ is our hypothesis;

• since $z\in\alpha$, we know that $z\subset\alpha$, hence $y\in z$ yields $y\in\alpha$;

• similarly, $x\in \alpha$.

Now, we have $x\in y\in z$ in $\alpha$. But $(\alpha,\in)$ is a well ordering and particularly $\in$ is transitive. Thus, $x\in y\in z$ implies $x\in z$, as desired.

• In my opinion, to prove that $z\in \alpha$ is transitive, we should start with the assumption $y\in z$ and get $y\subseteq z$ rather then assume that $x\in y$ and get $x\subseteq z$. May I ask how does this prove actually shows if $y\in z$ then $y\subseteq z$? – PropositionX Mar 21 '17 at 12:36
• Exactly! Thanks so much! – PropositionX Mar 21 '17 at 12:38
• Sorry but I think this time I am not given that a well-ordering is transitive in this version of definition, which I forgot. So without the transitivity of well ordering, could you please explain in another way using the definition above? – PropositionX Mar 21 '17 at 13:33
• Or is it possible to prove the transitivity of $\in$? – PropositionX Mar 21 '17 at 13:43
• Could you please tell me from which reference are you taking your definitions? I ask this because usually a well order in $P$ is a partial ordering $\leq$ such that for every $\emptyset\ne A\subset P$ there exists $\operatorname{min}⁡⁡A$. But a partial order is particularly transitive. Anyway, you may find this and this useful. – Renan Maneli Mezabarba Mar 21 '17 at 17:25

We must be careful to distinguish between a transitive set and a transitive binary relation on a set.

(1). Let $R$ be a binary relation on a set $X$. Then $R$ is a well-ordering iff

(1a). For all $x\in X$ we have $\neg (xRx)$, and also

(1b). For every non-empty $Y\subset X$ there is a unique $p\in Y$ such that $pRq$ for all $q\in Y$ \ $\{p\}.$....... We call such $p$ the unique $R$-least member of $Y.$

It follows that

(i) R satisfies Trichotomy, and

(ii) R is a transitive relation on $X$.

To prove (i): If $p,q\in X$ and $p\ne q$ then $\{p,q\}$ has a unique $R$-least member. So by (1b) we have $pRq$ or $qRp$, but not both of these..... And we have $\forall p\in X(\;\neg pRp)$ by (1a).

To prove (ii): If $pRq$ and $qRr$ then $\{p,q,r\}$ has a unique $R$-least member....

It can't be $r$ because by (i), $\;qRr\implies (\;q\ne r\land \neg rRq\;)$.

It can't be $q$ because by (i), $\;pRq\implies (\;p\ne q\land \neg qRp \;)$.

So $p$ is $R$-least. And we must have $p\ne r$, because $(p=r\land pRq\land qRr)\implies (pRq \land qRp)$, and we would violate (i) (Trichotomy). Since $p$ is the $R$-least member of $\{p,q,r\}$ and $p\ne r$, we have $pRr$.

(2). If $<$ is a well-ordering on a set $X$ then for any $Y\subset X$, the restriction of $<$ to $Y$ is a well-ordering of $Y$. For if $\phi\ne Z\subset Y$, then $\phi \ne Z\subset X.$ And any $Z$ that satisfies $\phi \ne Z\subset X$ has a $<$-least member.

(3). Let $A$ be an ordinal and $z\in A$.

(3a). $z\subset A$, so the restriction of $\in$ to $z$ is a well-order on $z$.

(3b). Now $\in$ is a well-order on $A$ and therefore $\in$ is a transitive relation on $A$. So by transitivity of the relation $\in$ on $A$,for any $y\in Z$ and any $x\in y$ we have $$(x\in y\land y\in z)\implies x\in z.$$ So for any $y\in z$ we have $y\subset z.$

By (3a) and (3b), $z$ is an ordinal.

• We cannot show that a member of an ordinal is an ordinal without using both parts of the def'n of an ordinal. A member of a transitive set need not be a transitive set. For example if $x=\phi , y=\{x\}, z=\{y\}$ then $A=\{x,y,z\}$ is a transitive set and $z\in A$ but $z$ is not a transitive set – DanielWainfleet Mar 22 '17 at 10:26

Let $\alpha$ be an ordinal and $z\in\alpha$. Thus $z\subsetneq\alpha$.

1. $z$ is well-ordered under $\in$. For $X\subseteq z,X\subsetneq\alpha$, which is well-ordered. Thus $z$ is well-ordered too.

2. $z$ is transitive. For all $m\in n\in z\in\alpha$, $m\in n\in z\subsetneq\alpha$, hence $m\in n\in\alpha$. Thus $m\in n \subsetneq\alpha$, hence $m\in\alpha$. Since $m,n,z\in \alpha$ and $m\in n\in z\in\alpha$, then $m\in z$. To sum up, $(m\in n\implies m\in z)\implies n\subsetneq z$.