How to prove that, if a is an ordinal and b is in a, then b is an ordinal?

Here are the definitions I'm using.

A set is an ordinal number if it is transitive and well-ordered by ∈.

A set T is transitive if every element of T is a subset of T.

My difficulty is mainly that I can't prove b is transitive. What's the magic?


Suppose $\beta\in\alpha$ and $\alpha$ is an ordinal. Let $\gamma\in\beta$. We need to prove that $\gamma\subseteq\beta$. Accordingly, let $\delta\in\gamma$. We need to show that $\delta\in\beta$.

Since $\alpha$ is an ordinal, and $\beta\in\alpha$, then $\beta\subseteq\alpha$. Therefore, $\gamma\in\alpha$. Again, this gives us that $\gamma\subseteq\alpha$. Thus $\delta\in\alpha$.

Now consider the set $\{\beta,\delta\}$. Since $\alpha$ is well-ordered by $\in$, we have that one of the following holds: $\beta=\delta$, $\beta\in\delta$, $\delta\in\beta$. We need to rule out the first two options.

If $\beta=\delta$, consider the set $\{\beta,\gamma\}$ and note that it contradicts well-foundedness: $\gamma\in\beta$ and $\beta=\delta\in\gamma$.

If $\beta\in\delta$, consider the set $\{\beta,\gamma,\delta\}$ and note that $\beta\in\delta\in\gamma\in\beta$, again contradicting well-foundedness.

The only option we have left is that $\delta\in\beta$, as we wanted.

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  • $\begingroup$ I see. I was missing the part "well-ordered by ∈", because in Jech's Set theory this is the first occurrence. Before this he only discusses well-order by <. $\endgroup$ – Zirui Wang Feb 12 '11 at 9:54
  • $\begingroup$ Even in the absence of the Axiom of Foundation, the cases $\beta=\delta$ and $\beta\in \delta$ contradict the linearity of the $\in$-order on $\alpha$.............+1 $\endgroup$ – DanielWainfleet Aug 7 '17 at 1:36
  • $\begingroup$ You said that ''we need to show that $\gamma\in\beta$'' but you showed that $\gamma\in\alpha$ ? $\endgroup$ – James Ensor Jun 18 '19 at 0:38
  • $\begingroup$ No, I didn't say that. I said we need to prove that $\gamma\subseteq\beta$", and then proceeded to prove just that. $\endgroup$ – Andrés E. Caicedo Jun 18 '19 at 1:14

Simpler proof than the one given:

Suppose $\beta\in\alpha$ and $\alpha$ is an ordinal. Suppose that $\gamma\in\beta$ and that $\delta\in\gamma$.

Since $\alpha$ is an ordinal, $\alpha$ is a transitive set. Therefore, since $\beta\in\alpha$ and $\gamma\in\beta$, we must have $\gamma\in\alpha$. Again, since $\gamma\in\alpha$ and $\delta\in\gamma$, $\delta\in\alpha$.

Now we have all the facts we need:

  1. $\in$ is a well-order on $\alpha$, which means that $\in$ is transitive on $\alpha$.

  2. $\beta,\gamma,\delta\in\alpha$

  3. $\gamma\in\beta$ and $\delta\in\gamma$

Since $\in$ is transitive on $\alpha$ and $\delta\in\gamma$ and $\gamma\in\beta$ and $\beta,\gamma,\delta\in\alpha$, we must have $\delta\in\beta$. Thus, $\beta$ is a transitive set. $\Box$

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  • 1
    $\begingroup$ Insufficient justification for the conclusion $\delta\in\beta$. $\endgroup$ – Zirui Wang Feb 26 '16 at 18:11
  • $\begingroup$ @ZiruiWang.Why? A well-order is necessarily a linear order, so $\in$ is a transitive binary relation on $\alpha.$ $\endgroup$ – DanielWainfleet Aug 7 '17 at 1:40

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