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I'm trying to establish the equivalence of the follow two assertions (Exercise 1.6 in page 52 of Set Theory and the Continuum Problem by Smullyan and Fitting).

AC. If $A$ is a non-empty set of non-empty sets, there is a function $f$ with domain $A$ such that for any $x \in A$, $f(x) \in A$.

Th. For every non-empty set $A$ there is a function $f$ such that for every proper subset $x$ of $A$, $f(x) \in x$.

I had no problem proving AC $\Rightarrow$ Th. For the reverse implication, however, I'm stuck. Here's my reasoning.

Claim: Th implies AC.

Incomplete proof. Let $A$ be a non-empty set of non-empty sets and consider the union set $\bigcup A$. Since $A$ is non-empty, so is $\bigcup A$, thus by Th there is a function $f$ such that for all proper subset $x$ of $\bigcup A$, $f(x) \in x$.

Let $y \in A$. Then $y \subseteq \bigcup A$, so if $y$ is a proper subset then $f(y) \in y$ as required. It remains to show that $y$ is a proper subset of $\bigcup A$. $\square$

I can't figure how to show that $y$ is a proper subset of $\bigcup A$. I've tried assuming $\bigcup A \subseteq y$, hoping to reach a contradiction, but the furthest I got in this direction is concluding that in this case, for any $x \in A$, $x \subseteq y$. Any help is appreciated.

purpleflyer
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    The statement of Th is incorrect. It is inconsistent as stated, so you cannot prove AC $\implies$ Th. Rather than "for every proper subset", it should say "for every non-empty subset". After all, $\varnothing$ is a proper subset of $A$, and it is impossible to have $f(\varnothing) \in \varnothing$. – Alex Kruckman Sep 03 '24 at 17:14
  • @AlexKruckman you're right obviously. I even unconsciously acknowledged it in my proof of AC $\Longrightarrow$ Th, because there I applied AC to the set $P = \mathcal P(A) \setminus \emptyset$.

    But checking the book, the statement of Th really says for every proper subset instead of non-empty subset. It seems changing the statement of Th to non-empty makes the proof work both ways, so that must have been the intended statement.

     – purpleflyer Sep 03 '24 at 19:11
  • NB I submitted an edit to this question which is obviously wrong (as I realized seconds afterwards) but apparently I have no way to retract it :( Sorry. – Jean Abou Samra Sep 03 '24 at 20:46
  • Do the authors define "proper" anywhere? It's rare but not unheard of for people to use the convention that $\emptyset$ is not a proper subset of any set. I assume that is what's going on here. See eg this – Izaak van Dongen Sep 09 '24 at 11:42
  • @IzaakvanDongen the authors seem to adopt the usual notion of proper subset; first they define proper informally on p. 7 by saying "(...) proper subset of $B$--that is, with a subset of $B$ which is not all of $B$ (it leaves out at least one element of $B$)." then on p. 37: "we shall use the notation $x \subset y$ to mean that $x$ is a proper subset of $y$ ($x \subseteq y$ but not $y \subseteq x$). – purpleflyer Sep 10 '24 at 18:18
  • Another amazing equivalence is the Axiom of Choice is equivalent to the Tychonoff's theorem – Ataulfo Sep 12 '24 at 01:52
  • Fair enough @purpleflyer! That is weird. Another way to fix it so Th does become equivalent to AC is to replace "$f(x) \in x$" by "$f(x) \in A \setminus x$". – Izaak van Dongen Sep 12 '24 at 11:09

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