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I was reading the book Sets, Models and Proofs and I'm now stuck at the part where they prove that Zorn's Lemma implies AC. In particular, this part:

"Proof. We assume that Zorn’s Lemma is true. Suppose given a surjective function $f : X → Y$. A partial section of $f$ is a pair $(A, u)$ where $A$ is a subset of $Y$ and $u : A → X$ a function such that $f (u(y)) = y$ for each $y ∈ A$. Given two such partial sections $(A, u)$ and $(B, v)$, put $(A, u) ≤ (B, v)$ iff $A ⊆ B $ and u is the restriction of v to A. Let P be the set of partial sections (A, u) of f ; then with the relation ≤, P is a poset, as is easy to see.

P is nonempty; this is left to you. . ."

What I'm confused about is this: Doesn't the existence of (A, u) require the Axiom of Choice? Because you have to choose an u for which f(u(y)) = y. For each y ∈ A, you have to pick an u(y) = x. I honestly don't see how I can prove that P is nonempty without using the Axiom of Choice or by using Zorn's Lemma, which is defined in the book as:

"Definition 1.3.4 Zorn’s Lemma is the following assertion: if (P , ≤) is a poset with the property that every chain in P has an upper bound in P, then P has a maximal element. Note that if P satisfies the hypothesis of Zorn’s Lemma, then P is nonempty. This is so because the empty subset of P is always a chain. However, checking that every chain has an upper bound in P usually involves checking this for the empty chain separately; that is, checking that P is nonempty."

Any help would be much appreciated.

Edit: For anyone that comes here after me, Do We Need the Axiom of Choice for Finite Sets? was also really helpful.

bowie schoonderbeek
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  • Welcome to MSE! Please use MathJax! As written your post is unreadable. Putting $ signs in between math formulas may help. – Vivaan Daga Mar 04 '23 at 12:56
  • Let $a\in X$ and $b=f(a)$. Then $(A,u)\in P$ if $A={b}$ and $u(b)=a$. This shows that $P\neq\varnothing$ and we did not use AC. – drhab Mar 04 '23 at 12:59

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No, in the proof, the author is just saying what means that a pair $(A, u)$ is a partial section of $f$, it doesn't say anything about if such a pair exists for any given $A\subseteq Y$, the important thing is that the set of ALL partial sections of $f$ is not empty, but this is easy to check, just take any $y\in Y$, since $f$ is surjective there exists $x\in X$ such that $f(x)= y$ and if you define $u : \{y\} \rightarrow X$ as $u(y):=x$ then the pair $(\{y\}, u)$ is a partial section of $f$.

Jorge S.
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  • But doesn't this mean you still just choose an x so that $f(x) = y$ for any $( {y} , u)$ ? Because for every $y∈Y$ there might be multiple x for which $f(x) = y$. – bowie schoonderbeek Mar 04 '23 at 13:27
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    @bowieschoonderbeek It is no problem to make a single choice or finitely many choices. The axiom of choice is only needed if we want to make infinitely many choices. Here you choose a single $y \in Y$ and then a single $x \in f^{-1}(y)$. – Paul Frost Mar 04 '23 at 13:39