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On Page 64, Set Theory, Jech(2006), define the following, by transfinite induction:

  • $V_0=\emptyset$,
  • $V_{\alpha+1}=P(V_{\alpha})$,
  • $V_{\alpha}=\bigcup_{\beta<\alpha}V_\beta$, if $\alpha$ is a limit ordinal.

How can we prove $V_{\alpha}$ is transitive by induction.

I tried what if other set, say $\{\{\emptyset\}\}$, is disignated as $V_0$. It turns out the transitive property fails. So somehow I should incoorperate $V_0=\emptyset$ into the induction. But then I stared at it, I stared at it, I just don't know what should I do.

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@Asafkaragila Sorry for incurring your inconvenience. I hope it is fixed now. –  Metta World Peace Dec 3 '12 at 18:38
Much obliged! Thanks. –  Asaf Karagila Dec 3 '12 at 18:41

2 Answers 2

up vote 2 down vote accepted

First let us prove the following claim.

Claim I: If $x$ is transitive then $\cal P(x)$ is transitive.

Proof. Suppose that $z\in y\in\cal P(x)$, then $y\subseteq x$, and $z\in x$. However $x$ is transitive so $z\subseteq x$ as well. Therefore $z\in\cal P(x)$ as wanted. $\square$

Now we will prove this second claim as well.

Claim II: If $\mathcal X=\{x_i\mid i\in I\}$ is a $\subseteq$-chain of transitive sets then $\bigcup\cal X$ is transitive.

Proof. Let $x\in\bigcup\cal X$, then for some $i\in I$ we have that $x\in x_i$. By transitivity of $x_i$ we have that $x\subseteq x_i$ and therefore $x\subseteq\bigcup\cal X$, as wanted. $\square$

Now you can use Claim I for the successor steps and Claim II for limits, and of course fire off the induction with the proof that $\varnothing$ is transitive.

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Thank you for your answer with explicit use of two equvelents of transitivity. –  Metta World Peace Dec 3 '12 at 19:35

If $V_{\alpha}$ is transitive, consider $x \in y \in P(V_{\alpha})$. Then $y$ is a subset of $V_\alpha$, so that $x$ is in a subset of $V_{\alpha}$ and hence is an element of $V_{\alpha}$, so must also be an element of $P(V_{\alpha})$. Thus $P(V_{\alpha}) = V_{\alpha+1}$ is transitive.

If $V_{\beta}$ are transitive for $\beta < \lambda$, then $\bigcup_{\beta < \lambda} V_{\beta} = V_{\lambda}$ is transitive because if $x \in y \in V_{\lambda}$ then $x \in y \in V_{\beta'}$ for some $\beta' < \lambda$, and since $V_{\beta'}$ was transitive, $x \in V_{\beta'}$, hence $x \in V_{\lambda}$.

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Thank you for your answer. –  Metta World Peace Dec 3 '12 at 19:36

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