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Concerning the above proof of how to show that AC implies Zorn's lemma, there is just one thing that I don't get.

Namely, they say that the definition of $F$ implies immediately $(1)$, but I don't get why $F(\alpha)

Here are the references: http://www.math.uni-bonn.de/ag/logik/teaching/2016WS/Aktuelles_Skript.pdf

(Pages 28 & 29)

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Each time we constructed some chain, and if possible we choose a strict upper bound. Namely, an upper bound which is strictly larger than all the elements of the chain. But the chain is made of the previously-chosen elements.

So if $\alpha<\beta$, then $F(\beta)$ is chosen as an element which is strictly larger than $F(\alpha)$ by the very definition of $F(\beta)$.

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    But if $F(\beta) \neq P$, according to the upper definition, does the author refer to $F(\beta)$ as $F(\beta) = p$ for some p chosen with the property that it is strict upper bound not contained in $F[\beta]$ or $F(\beta)=g(p)$ with $p$ having the same properties as just mentioned? – 2017-02-20
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    I don't understand your question. – 2017-02-20
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    I am asking this just to make clear in my head the definition of $F$. When he defines $F$ for $\alpha$ such that $F(\alpha) \neq P$, is he saying that, in that case $F(\alpha)$ is equal to $p$ or $F(\alpha)=g(p)$ – 2017-02-20
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    No. It does not say that. – 2017-02-20
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    So $F(\alpha)=g(p)$, right? – 2017-02-20
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    What is $p$ here? $g$ is a choice function from subsets of $P$. I don't know what $p$ is, and why $g(p)$ is even defined. And in the case that you mean $g(P)$, how would that make any meaningful definition? Since $g$ is a function, $g(P)$ will always have the same value, regardless to how many times you tried to compute it. – 2017-02-20
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    But $p$ is a strict upper bound (contained in $P$ \ $F[\alpha]$) which is why it made me think of $F(\alpha)$ as it is equal to the precise upper bound $p$ which $g$ chooses out of all upper bounds not contained in $F[\alpha]$. – 2017-02-20
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    Yes, $p$ is **an** upper bound. $g$ chooses *from the set of strict upper bounds*. So $p$ is the *result* of $g$ choosing from some non-empty set of strict upper bounds. – 2017-02-20