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I am having difficulty understanding why the following paragraph, copied from "Foundations of Set Theory, by Fraenkel, Bar-Hillel & Levy", Section 4.1, concludes that the subsets using the axiom of separation are not uniquely determined. The wording looks to have $y \in y$, so I'm guessing there are some typos, but I cant work out what they are or how the "not uniquely determined" arises:

"In our axiomatic theory, this way of introducing the subset $c$ of $∪ t$ is not in accordance with the axiom of subsets – except for the trivial case that every member of $t$ contains one member only, in which case $c = ∪ t$ satisfies our condition. In the general case, the subset $c$ of $∪ t$ has not been defined by a definite condition $B(x)$ that is characteristic, among all $x ∈ ∪ t$, of the $x ∈ c$ and only of them. On the contrary, suppose $c ⊆ ∪ t$ is of the desired kind and $y ∈ c$ belongs to a certain $y ∈ t$; then, replacing $y$ by a different member $y′$ of the same $y ∈ t$ will yield a new subset $c′ ⊆ ∪ t$ which differs from $c$, while $c′$ is also a subset of $∪ t$ with the desired property. Thus, contrary to the subsets postulated by the axiom of subsets, the subsets of $∪ t$ needed for our purpose are not uniquely determined."

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    $\begingroup$ It rather concludes that the subset whose existence is stated by the axiom of choice, is not arising from the axiom of separation: the subset might not be defined by a first order formula. $\endgroup$
    – Berci
    Commented Dec 20, 2019 at 16:14
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    $\begingroup$ I’m not sure where you’re seeing $y\in y$. Also, please use mathJax so it is easier to read. $\endgroup$ Commented Dec 20, 2019 at 16:26
  • $\begingroup$ @spaceisdarkgreen. OK on the mathJax. The y∈y arises indirectly from the words : y ∈ c belongs to a certain y ∈ t. $\endgroup$
    – user239186
    Commented Dec 21, 2019 at 10:18

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This is something I learned from Azriel Levy when we were teaching set theory together back in Jerusalem.1

He pointed out to me this main difference between the Axiom of Choice and the rest of the axioms of ZF: all the axioms that assert existence (i.e., all but Extensionality) tell you that from a given set $x$ you can generate a set $y$.

The Axiom of Choice, however, does not tell you how to generate the set it guarantees to exist. It just tells you of its existence.

If we consider the simplest version of the Axiom of Choice, from a formal set theoretic perspective, that would be "If $t$ is a set of pairwise disjoint sets, there is another set $c\subseteq\bigcup t$ such that for each $y$ there is a unique $x\in c\cap y$."

The point is that $c$ need not have a concrete definition. Of course, it might, for example if each $y\in t$ has a single element, then $c=\bigcup t$. Or if every $y$ in $t$ has the form $\{x,\{x\}\}$ for some $x$, then we can take $c$ to be the set of $\in$-minimum from each $y$ (recall that we assumed these are pairwise disjoint).

But if, for example, $t=\{x\times\{x\}\mid x\subseteq\mathcal P(\omega)\}\setminus\{\varnothing\}$, then there is no way to specify such selector $c$.

Often in the first few lectures about the Axiom of Choice each year (yes, we taught together for three years2) we would have at least one, and probably more, person asking why the Axiom of Choice does not follow from Subset/Bounded Comprehension/Substitution (call it what you will), since it seems to be a way to generate subsets. To which the standard reply is "Okay, how do you define this choice?" which often flusters the students, and then they understand the point.


  1. Yes, I just namedropped Azriel Levy.

  2. Yes, I am bragging.

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    $\begingroup$ I feel like you probably knew this before you were teaching set theory with Levy. $\endgroup$ Commented Dec 20, 2019 at 20:22
  • $\begingroup$ Yes, but it wasn't as explicit in my mind as after we had that discussion. $\endgroup$
    – Asaf Karagila
    Commented Dec 20, 2019 at 20:23
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    $\begingroup$ But can you namedrop Azriel's name in casual conversation? That's the real treasure of having worked with him! :-P $\endgroup$
    – Asaf Karagila
    Commented Dec 20, 2019 at 20:30
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    $\begingroup$ @LittleCheese: No, as I said, this very much depends on the set $t$. But there is no formula that works for every set $t$. And indeed, it is possible that there is a set for which no formula works to give a selector $c$ out of $\bigcup t$ (even if AC holds!) $\endgroup$
    – Asaf Karagila
    Commented Dec 20, 2019 at 22:15
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    $\begingroup$ I would say the Power-Set Axiom also does not tell you how to generate a Power Set, but merely declares it existence. But AC is always the one that students have difficulty with. $\endgroup$ Commented Dec 29, 2019 at 23:37

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