# Closure of a transitive set under Gödel operations is transitive: why?

I've a small question.

If I have $X$ a transitive set, why is its closure under Gödel operations still transitive?

Kind thanks,

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I will use the Gödel operations as defined in Jech's Set Theory (millennium edition) in Part II, chapter 13, page 178.

Let's assume that $X$ is a transitive set. First let $\operatorname{cl}(X) = \bigcup_{n\in\omega} X_n$, where $X_0=X$ is $X_{n+1}=X_n\cup\{G_i(y_1,y_2,y_3): i\in 10\land y_1,y_2,y_3\in X_n\}$. Now assume that $z\in y\in \operatorname{cl}(X)$, and let's assume that $y\in X_n$. If $n=0$ we have $z\in \operatorname{cl}(X)$. Proceed with induction: If $y\in X_{n+1}\setminus X_n$, $y$ is the output of one of the Godel operations. If it's $G_1$, then $y=\{x_1,x_2\}$. By definition, $x_1, x_2$ are elements of the closure. If it's $G_2$, then $y=x+1\times x_2$. Its elements are of the form $\{\{a\},\{a,b\}\}$, where $a\in x_1$ and $b\in x_2$, which by the induction hypothesis implies that $a,b\in \operatorname{cl}(X)$. Hence since $\{a\}=G_1(a)$ and $\{a,b\}\in G_1(a,b)$ and $(a,b)=G_1(\{a\},\{a,b\})$, we have that $(a,b)\in \operatorname{cl}(X)$. The rest of the operations are easy: Most of them are subsets of some Cartesian product, which I just show that works, or set operations, which directly give transitivity.

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Thanks Apostolos. –  Girdle Dec 6 '11 at 0:14