Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've a small question.

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

Kind thanks,

share|improve this question
add comment

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer
    
Thanks Apostolos. –  Girdle Dec 6 '11 at 0:14
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.