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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 ?

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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.

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  • $\begingroup$ 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

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