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.


1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$
    – Andrew Li
    Mar 19 at 4:22
  • $\begingroup$ That is exactly the case the term "at most one" is used to account for. However, the above answer demonstrates that that is not the only way to account for the case, by providing another way to do so. $\endgroup$ Mar 19 at 4:29
  • $\begingroup$ @AndrewLi Yes. Gaurav is exactly right. $\endgroup$ Mar 19 at 5:26

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