I'm currently reading "Axiomatic Set Theory" by Suppes, and the book gives a proof of the existence of the intersection (which relies on the Axiom of Separation). While I understand the idea of the proof, I don't understand why it is the Theorem is stated as follows:

$$\forall x \exists y(x\in y \Leftrightarrow x\in A \wedge x\in B)$$

By the definition of the Axiom of Separation don't we need a "$z$" such that $x\in z$ (?)

The formula would then be:

$$\forall x \exists y (x \in y \Leftrightarrow x\in z\wedge(x\in A\wedge x\in B))$$

Why is the proof not given this way? Also, in the same way that a separate "Axiom of Union" is introduced for the union, is the above method of "defining" the intersection really the best way to do it?

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    $\begingroup$ Please learn latex. $\endgroup$ – Rene Schipperus Sep 28 '14 at 2:20
  • $\begingroup$ Yes $z=A$ see your previous question. $\endgroup$ – Rene Schipperus Sep 28 '14 at 2:40
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    $\begingroup$ I don't have that book. Is it really $\forall x\exists y$ and not $\exists y\forall x$ in the book? $\endgroup$ – bof Sep 28 '14 at 2:42

Now, I have been reading the book. I summarize

Axiom 1 (of Separation). Let $A$ a set, and for each $x \in A$, let $\varphi(x)$ a property pertaining to $x$. Then there exists a set $C := \{x \in A : \varphi(x) \text{ is true} \}$ (or $\{x \in A : \varphi(x)\}$ for short), whose elements are precisely the elements $x$ in $A$ for which $\varphi(x)$ is true. $$\exists C \, \forall x \; (x \in C \iff x \in A \;\land\; \varphi(x) ).$$

Axiom 2 (of Extensionality). Two sets $A$ and $B$ are equal, $A = B$, if every element $x$ of $A$ belongs also to $B$, and every element $y$ of $B$ belongs also to $A$. $$\forall x \; (x \in A \iff x \in B) \implies A = B.$$

Your theorem

Theorem 3. Let $A$ and $B$ be sets. Then exists a unique set $C$ whose elements belongs to both $A$ and $B$. $$\exists ! C \, \forall x \; (x \in C \iff x \in A \;\land\; x \in B).$$

Now, to prove the theorem we need show the existence and uniqueness. First the existence. Let $A$ and $B$ be sets, and let $\varphi(x) := x \in B$. Then, by Axiom 1, we have $$\exists C \, \forall x \; (x \in C \iff x \in A \;\land\; \varphi(x) ),$$ i.e., $$\exists C \, \forall x \; (x \in C \iff x \in A \;\land\; x \in B ).$$ This prove the existence of the intersection set $C$ for any sets $A, B$.

We now show the uniqueness. Let $A$ and $B$ sets. Suppose there exists two sets $C, C'$ such that $$\exists C \, \forall x \; (x \in C \iff x \in A \;\land\; x \in B )$$ and $$\exists C' \, \forall x \; (x \in C' \iff x \in A \;\land\; x \in B ).$$ Using the notation, we have $$x \in A \;\land\; x \in B \iff x \in C',$$ i.e., $$x\in C \iff x \in A \;\land\; x \in B \iff x \in C'.$$ This means, the statements $x\in C$, $x \in A \;\land\; x \in B$, and $x \in C'$ are equivalents. Thus, we have $$x \in C \iff x \in C'.$$ By Axiom 2, we have $C = C'$ as desired. $\;\Box$

Now, note that to use the Axiom 1, you need two sets: a reference set from which construct the intersection, and another to define the property. Also, another valid definition of intersection is $$A \cap B = \{x \in A : x \in B\}.$$ So the set $z$ that you mention is $A$, and your property $x \in A \; \land \; x \in B$ should be just $x \in B$.

Finally, in Set theory, the Axiom of union exists because this does not follow from the other axioms (empty set, extensionaity, separation, etc). These axioms allow us to build smaller sets from other reference sets. But the union set is a larger set from anothers, because of that this axiom is necessary.

  • $\begingroup$ @AF3: Maybe you could add tags:[ self-learning] and [proof-explanation]. $\endgroup$ – Cristhian Gz Sep 28 '14 at 14:51
  • $\begingroup$ Thanks, Christian. I guess the point is that the Axiom of Separation may "separate" off subsets of a given set, and that we will not ever "get out" of an existing set A through the axiom of separation. However, we COULD form a proper subset of A through the "union" of (finitely many) subsets of A using the Axiom of Separation, right? We would take P(x) to be the disjunction of x's membership in some (finitely many) subsets of our "referenceset" A, and this should be covered under the Axiom. $\endgroup$ – AF3 Sep 28 '14 at 15:45
  • $\begingroup$ Exactly. In the book, the empty set is the one whose existence is supoosed. Then you will see how to form new sets form the axioms of pair and union (from the empty set) and, by axiom of separation (or specification) you form the mainly sets: natural numbers, cardinals, ordinals, etc. Wishes for reading. $\endgroup$ – Cristhian Gz Sep 28 '14 at 16:01
  • $\begingroup$ Great, thanks. So I guess the plan is like...we're going to use the Union Axiom and Power Set Axiom to form the Cartesian Product, and that once we have the Cartesian Product, then we can get to all of the "familiar" ideas like function, relations and so on...Then we can start building up the numbers through a combination of these axioms and their consequences...and from building up the numbers we get...lots of other stuff (since functions will just be various subsets of the real numbers which we can form through things like The Axiom of Separation)? $\endgroup$ – AF3 Sep 28 '14 at 16:23

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