Tarski-Grothendieck set theory, the axiom of pairing and the axiom of specification - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T16:20:56Z http://mathoverflow.net/feeds/question/104464 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104464/tarski-grothendieck-set-theory-the-axiom-of-pairing-and-the-axiom-of-specificati Tarski-Grothendieck set theory, the axiom of pairing and the axiom of specification Gérard Lang 2012-08-11T08:46:56Z 2012-08-11T15:30:49Z <p>I am building upon MO question 102846 concerning the Tarski-Grothendieck set theory (TG). I have two questions; 1/ I think that it is possible that the axiom of pairing (axiom 4 of the TG theory presented in question 102846) could be deleted. So, my question is: Is axiom 4 independant of the others axioms of the TG set theory ? 2/ In the considered question, it is asserted that the axiom of specification is a consequence of Tarski"s axiom A. My question is: what are the others axioms needed to have this result and could I have a proof ? Gérard Lang</p> http://mathoverflow.net/questions/104464/tarski-grothendieck-set-theory-the-axiom-of-pairing-and-the-axiom-of-specificati/104497#104497 Answer by Trevor Wilson for Tarski-Grothendieck set theory, the axiom of pairing and the axiom of specification Trevor Wilson 2012-08-11T15:30:08Z 2012-08-11T15:30:08Z <p>Assume all the axioms of TG except for pairing. We will show that pairing and specification (a.k.a. separation) both follow, with the caveat in the following paragraph.</p> <p>We will also assume that the empty set exists. This does not seem to follow from the TG axioms as defined in the linked question <a href="http://mathoverflow.net/questions/102846/" rel="nofollow">http://mathoverflow.net/questions/102846/</a> or on Wikipedia <a href="https://en.wikipedia.org/wiki/Tarski-Grothendieck_set_theory" rel="nofollow">https://en.wikipedia.org/wiki/Tarski-Grothendieck_set_theory</a>, which do not seem to imply either "there is a set" or "if there is a set then there is an empty set." The natural way to proceed seems to be to postulate the existence of the empty set.</p> <p>(1) For pairing, suppose we are given sets $x$ and $y$ and we want to form the pair $\lbrace x,y \rbrace$. Let $A$ be a Tarski set containing $x$. Then $A$ also contains some element other than $x$ (this can be seen by considering separately the two cases $x = \emptyset$ and $x \ne \emptyset$.) Define a class function $F$ with domain $A$ by $$F(z) = \begin{cases} x &amp; \text{if z = x}\newline y &amp;\text{if z \in A and z \ne x}. \end{cases}$$ The range of $F$ is $\lbrace x,y \rbrace$, which is a set by replacement.</p> <p>(2) For specification, let $A$ be a set and $\varphi$ be a formula. We want to show that the class $B = \lbrace x \in A : \varphi(x)\rbrace$ is a set. If $B$ is empty then we are done because the empty set exists. So we may assume that there is $x_0 \in A$ such that $\varphi(x_0)$ holds. Then $B$ is the range of the class function $F$ with domain $A$ defined by $$F(x) = \begin{cases} x &amp; \text{if x \in A and \varphi(x)}\newline x_0 &amp;\text{if x \in A and \neg \varphi(x)}, \end{cases}$$ so it is a set by replacement. </p>