In Enderton's text , subset axioms is provided as an axiom of set theory ( it's axiom schema)

The axioms is :

For each formula $\tau$ not containing B as a symbol , the following is an axiom: $\forall t_1 ... \forall t_k \forall c \exists B \forall x (x \in B \iff x \in c \wedge \tau ) $

Now . given a set $A$ ( its elements - if any exists - are themselves sets ) , how can we define $\bigcap A$ ?

The author say that , for a fixed element $c$ in $A$ , $x \in \bigcap B \iff x\in c \wedge \text{x belongs to every other member of A}$

So here $\tau$ is '$\text{x belongs to every other member of A}$'

Of course , we can write 'x belongs to every other member of A' formally easily , but my question is, going back to the formal axiom , we notice that there is '$\forall c$' before $\exists B$ , but the author's definition of the intersection of $A$ has no '$\forall c$' before '$\exists B$' and to add it then we have to change the expression to assure that $c$ is an element of $A$ not any set in general, but we are not allowed to change any thing except in $\tau$. I don't know if changing $\tau$ is allowed or not , So How to do this ?

My attempt is to define the intersection as :

$\bigcap A = \{x|\forall A \exists B \forall x [x \in B \iff x\in c \wedge ( c \in A \wedge \forall a\in A (a\ne c \rightarrow x\in a))]\}$

(notice $c\in A$ inside $\tau$ , I put it to specify that $c$ must be an element of $A$ not any one)

Is this definition right ? if not , How to do that ?


I am not sure if I am addressing your problem, but you should distinguish two things: The definition of $\bigcap A$ and the proof that an object satisfying the definition actually exists.

First you want to define the meaning of $\bigcap A$. This is simply done by $$\bigcap A = \left\{x\colon \forall y(y\in A\rightarrow x\in y) \right\}, $$ which just means that $$x\in\bigcap A \iff \forall y(y\in A\rightarrow x\in y). $$ Next we have to show that in case that $A$ is non-empty there actually is such a set, i.e. a set $B$ such that $$x\in B \iff \forall y(y\in A\rightarrow x\in y). $$ Formally, you would want to derive $$\forall A((\exists c(c\in A))\rightarrow\exists B(\forall x(x\in B \leftrightarrow \forall y(y\in A\rightarrow x\in y)))) $$ from the axioms. I will sketch this proof informally.

If $A$ is non-empty, we can pick a $c\in A$ and note that actually

$$\forall y(y\in A\rightarrow x\in y) \iff (x\in c) \land \forall y(y\in A\rightarrow x\in y). $$ Now the axiom guarantees the existence of that $B$. Indeed it says that for every set $c$ there is a set $B$ such that

$$x\in B \iff (x\in c) \land \forall y(y\in A\rightarrow x\in y), $$ so in particular for the $c$ that we have chosen.

  • $\begingroup$ My question is , How to define this set in the form of the axiom ? How can we say that $c$ is an element of $A$ in form of the axiom ?(Could you see my definition in the queustion and see if it works plz ? ) $\endgroup$ – Element Oct 6 '13 at 13:22
  • $\begingroup$ @Element, you are not trying to do the right thing. You do not have to “define $\bigcap A$ in the form of the axiom”. I have expanded my answer a bit, maybe that helps. Your attempt at a definition does not make much sense. You have $\forall A$ and $\exists x$ in it, and the $A$ and $x$ to the right of them do not refer the same thing as the $A$ and $x$ to the left of them. Also, the $c$ appears out of nowhere without a $\forall$ or $\exists$. $\endgroup$ – Carsten S Oct 6 '13 at 13:57
  • $\begingroup$ thanx , I think I understand the issue now :) $\endgroup$ – Element Oct 6 '13 at 14:24

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.