2
$\begingroup$

Note: This question is not a duplicate of this, due to the corrected version of the book.

I'm self-studying Tao's Analysis I. In the corrected third edition, Tao promotes the what was earlier a definition of equality of sets, to the status of an axiom, after someone pointed out an issue with that.

Axiom 3.2 (Equality of sets) Two sets $A$ and $B$ are equal, iff every element of $A$ is an element of $B$ and vice versa. To put it another way $A=B$ if and only if every element $x$ of $A$ belongs also to $B$ and every element $y$ of $B$ belongs also to $A$.

Just after this (well, there's Example 3.1.4 in between), he mentions the following, without justification.

The "is an element of" relation $\in$ obeys the axiom of substitution (see section A.7).

I quote the relevant part of section A.7 here:

(Substitution axiom). Given any two objects $x$ and $y$ of the same type, if $x=y$, then $f(x) = f(y)$ for all functions or operations $f$.
Similarly, for any property $P(x)$ depending on $x$ if $x=y$, then $P(x)$ and $P(y)$ are equivalent statements.


Questions: Why does Axiom 3.2 ensure that $\in$ satisfies the axiom of substitution? This is what I have in mind: To prove that $\in$ satisfies axiom of substitution, one needs to first define an equality relation and $\in$-relation for sets. Then only we can proceed in any way, as far as I can see. But Axiom 3.2 does not define any of these relations. It just gives a condition that equality and $\in$ must satisfy.

Don't we need another axiom stating that there is an equality relation on sets, and a relation $\in$ on sets such that $\in$ obeys axiom of substitution with respect to this equality?

$\endgroup$
4
  • 1
    $\begingroup$ I'm not familiar with the axiomatic set-up used in this book (so this is a comment, not an answer), but in most modern treatments of axiomatic set theory, the concept of equality and the axiom of substitution are part of the underlying logic. The membership relation, $\in$, is the primitive notion of set theory. Neither $=$ nor $\in$ is defined in terms of anything more basic. $\endgroup$ Apr 24 '21 at 16:52
  • $\begingroup$ @AndreasBlass So we just assume that the equality that is talked about in the axiom of extensionality is indeed a 'valid equality' and the $\in$-relation is a 'valid relation' in the sense that this equality (that extensionality talks about) follows axioms of equality - reflexivity, symmetry, transitivity, and that the $\in$-relation obeys axiom of substitution w.r.t. this equality? $\endgroup$
    – Atom
    Apr 24 '21 at 18:02
  • $\begingroup$ Yes, in the usual systems it is assumed, as part of the underlying logic, that $=$ satisfies all the usual axioms of equality, including substitution for all the operations and predicates of the theory. The axiom of extensionality then gives additional information about the set-theoretic notion of membership, namely a certain connection with the logical notion of equality. $\endgroup$ Apr 24 '21 at 18:07
  • $\begingroup$ @AndreasBlass This makes sense. $\endgroup$
    – Atom
    Apr 25 '21 at 6:05

Your Answer

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

Browse other questions tagged or ask your own question.