I'm going through Enderton's Elements of Set Theory, as I heard it is a gentle introduction to set theory. I'm a little confused on how the subset axioms are used. In the text, the axiom is given as:

Subset Axioms For each formula $\underline{\hspace{1cm}}$ not containing $B$, the following is an axiom: $$ \forall t_1\cdots\forall t_k\forall c\exists B\forall x(x\in B\Leftrightarrow x\in c \wedge \underline{\hspace{1cm}}). $$

Later, the text asserts the existence of the set $\{A\cup X\ | \ X\in\mathscr{B}\}$, call it $\mathscr{D}$, saying that since $A\cup X\subseteq A\cup\bigcup\mathscr{B}$, $\mathscr{D}$ is a subset of $\mathscr{P}(A\cup\bigcup\mathscr{B})$. A subset axiom produces $$ \{t\in\mathscr{P}(A\cup\bigcup\mathscr{B})\ |\ t=A\cup X\ \text{for some}\ X\in\mathscr{B}\}. $$

My question is, what exactly is that subset axiom? Would it be something of the form $$ \forall t_1\cdots\forall t_k\forall c\exists B\forall x(x\in B\Leftrightarrow x\in c \wedge [(x=t_1\cup t_2)\vee(x=t_1\cup t_3)\vee\cdots\vee(x=t_1\cup t_k)]) $$

where we take $A$ to be a particular instance of $t_1$, $\mathscr{P}(A\cup\bigcup\mathscr{B})$ to be a particular instance of $c$, and $t_2,\dots,t_k$ to be particular instances of members of $\mathscr{B}$?

Or perhaps something like $$ \forall t_1\forall t_2\forall t_3\forall c\exists B\forall x(x\in B\Leftrightarrow x\in c \wedge [\exists t_3\in t_2(x=t_1\cup t_3)]) $$

where in this case $\mathscr{B}$ is a particular instance of $t_2$?, and then $\mathscr{D}$ would be the set $B$ which has been proven to exist. Thanks, right now I'm just not used to what should go in that blank spot for the formula.

  • $\begingroup$ Recall the Russell paradox (or any other "My set is too big" paradox), the consequences were that some collections are not sets. These were called "proper classes" (proper since sets are also classes). The subset and replacement axioms are to ensure that if you took a subcollection of a set - you will have a set (and in the case of replacement - if you have a function whose domain is a set then so is its image). $\endgroup$ – Asaf Karagila Dec 23 '10 at 6:59

If you want to form the set $$ \{t\in\mathscr{P}(A\cup\bigcup\mathscr{B})\ |\ t=A\cup X\ \text{for some}\ X\in\mathscr{B}\}. $$ then a natural instance of the axiom of separation (the "subset axiom") is $$ (\exists B)(\forall t) ( t \in B \Leftrightarrow t\in\mathscr{P}(A\cup\bigcup\mathscr{B}) \land (\exists X)[X \in \mathscr{B} \land (\forall s) (s \in t \Leftrightarrow s \in A \lor s \in X)]). $$ This formula is obtained by simply unpacking the original abbreviated formula, for example by expanding the "=" using its set-theoretic definition.

There is a general principle that if you have assumed a formula of the form $(\forall z)\phi(z)$ then, for any set $Y$, the formula $\phi(Y)$ will be true as well. This is called (universal) instantiation. The way you get to the separation axiom above is to start with $$ (\forall C_1)(\forall C_2)(\forall C)(\exists B)(\forall t) ( t \in B \Leftrightarrow t\in C \land (\exists X)[X \in C_1 \land (\forall s) (s \in t \Leftrightarrow s \in C_2 \lor s \in X)]). $$ and then instantiate it with $C_1 = \mathscr{B}$, $C_2 = A$, $C = \mathscr{P}(A\cup\bigcup \mathscr{B}) $.

  • $\begingroup$ Thanks, this was very helpful. $\endgroup$ – yunone Dec 23 '10 at 3:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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