# How is axiom of substitution satisfied for $\in$? (Tao' Analysis I)

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?

• 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. Apr 24 '21 at 16:52
• @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?
– Atom
Apr 24 '21 at 18:02
• 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. Apr 24 '21 at 18:07
• @AndreasBlass This makes sense.
– Atom
Apr 25 '21 at 6:05