# Question on Axiom of replacement

This axiom comes from chapter 3 "set theory" of Tao Analysis I

Axiom $$3.6$$ (Replacement). Let $$A$$ be a set. For any object $$x \in A$$ and any object $$y$$, suppose we have a statement $$P(x, y)$$ pertaining to $$x$$ and $$y$$, such that for each $$x\in A$$ there is at most one $$y$$ for which $$P(x,y)$$ is true. Then there exists a set $$\{y: P(x, y) \text{ is true for some } x \in A\}$$ such that for апy object $$z$$, $$z\in \{y : P(x, y)\text{ is true for some } x \in A\} \iff P(x,z)\text{ is true for some } x \in A.$$

For me, this axiom implies a way to construct a new set $$B$$ from the original set $$A$$, and it's a mapping from set $$A$$ to set $$B$$. So it's obvious that for each $$x\in A$$ there is exactly one $$y$$ for which $$P(x,y)$$ is true.

My question is: Can I replace "at most one" with "exactly one" in the axiom?

Thank you guys, I think my question should be closed. Because I got this from here:The axiom of replacement basically says that if A is a set and f is an operation on elements of A, then $$\{f(x) : x \in A\}$$ is a set. Here the operation f may return an undefined result (because for each $$x$$, the statement $$P(x,y)$$ is true for at most one $$y$$ rather than exactly one $$y$$). So to construct the set $$\{x \in A : P(x) \text{ is true}\}$$, we can define $$f(x)$$ to be $$x$$ if $$P(x)$$ is true, and leave $$f(x)$$ undefined if $$P(x)$$ is false.

Yes they are equivalent. Using the axiom of separation you can construct a subset $$\mathcal{A}\subseteq A$$, such that for each $$x\in \mathcal{A}$$ there is exactly one $$y$$ for which $$P(x,y)$$ is true, applying your version of replacement on this set, we can prove Tao’s version.
• I think "at most one" implies that maybe for example $x_1 \in A$ ,there is no $y$ corresponding to $x_1$. Is this the reason why Tao use "at most one" in the axiom? Mar 19 at 4:22