# If $\alpha$ is an ordinal set and $\beta\in\alpha$ then $\beta$ is an ordinal set also

In Jech's Set theory book it is stated that

If $\alpha$ is an ordinal set and $\beta\in\alpha$ then $\beta$ is an ordinal set also. Proof: this follow from the definition of ordinal set.

And the definition of ordinal set says

A set is ordinal if it is transitive and well-ordered by $\in$.

However I can't see how from the definition follows the first statement. I can see that if $\beta\in\alpha$ and $\gamma\in\beta$ then $\gamma\in\alpha$ because $\alpha$ is a transitive set, that is, $\beta\in\alpha\implies\beta\subset\alpha$.

Then $\gamma\subset \alpha$. But Im unable to see why $\beta$ must be also transitive, that is, why $\gamma\subset\beta$. Can someone clarify this point? Thank you.

EDIT: ok, seeing in wikipedia I saw that there are two different (but equivalent) definitions of transitivity. My definition is

A set $x$ is transitive if $y\in x\implies y\subset x$.

The other definition state that

A set $x$ is transitive if $y\in x\land z\in y\implies z\in x$.

Then using the second definition clearly $\beta$ is transitive (I guess that this is what @i707107 tried to point in the comments). However I'm using the first definition, what makes this not obvious. Then the question reduces to show the equivalence between the two definitions of transitivity.

• If $\beta$ is a subset of $\alpha$, the the order in $\alpha$ applies to every element of $\beta$. – Sungjin Kim Jun 12 '18 at 2:53
• @i707107 yes, I know, but the order is stated by $\in$ not by $\subset$ – Masacroso Jun 12 '18 at 2:55
• Every element of $\alpha$ is ordered by $\in$, so every element of $\beta$ is also ordered by $\in$. – Sungjin Kim Jun 12 '18 at 2:57
• The two definitions are literally the same. The first one uses the shorthand for the definition of $\subseteq$, whereas the second writes it explicitly in terms of $\in$. There is nothing to prove. Let that be a lesson to you, to always unwind definitions when you're stuck. – Asaf Karagila Jun 12 '18 at 7:26

To see equivalence of the two definitions, first suppose $y \in x \wedge z\in y \Rightarrow z \in x$. Now $y \subset x$ iff $z \in y \Rightarrow z \in x$. So if $y \in x$ (what we must suppose to prove the statement $y \in x \Rightarrow y \subset x$) and $z \in y$ (which we must suppose to prove the result of the conditional), we have exactly the antecedent $y \in x \wedge z\in y$ of our supposition, so $z\in x$, so $y \subset x$, and we conclude that $y\in x \Rightarrow y\subset x$.
To prove the other side, suppose that $y\in x \Rightarrow y \subset x$, and suppose that $y\in x \wedge z\in y$. Now $y\in x \wedge z\in y \Rightarrow y \in x$, obviously, and by our first supposition, $y\in x \Rightarrow y \subset x$. So all together, we have $y\in x \wedge z\in y \Rightarrow y \in x \Rightarrow y \subset x$, and we are done.
Ok, I solved it: suppose that $\gamma\not\subset\beta$. Then there is some $\delta\in\gamma$ such that $\delta\notin \beta$.
However we knows (by the first definition of transitive set) that $\gamma\subset\alpha$ so $\delta\in\alpha$. Hence, because $\alpha$ is well-ordered by $\in$ and $\delta,\beta\in \alpha$ if $\delta\notin\beta$ then necessarily $\beta\in \delta$, but then by the thansitivity of an order relation we find that $\delta\in\gamma\in\beta\in\delta$, so $\delta\in\delta$, what is not possible because order relations in Jech book are defined by the axiom $x\not <x$, in this case $x\notin x$.
Hence $\gamma\subset\beta$ and we can conclude that $\beta$ is a transitive set, so it is also an ordinal.