Equivalences on the Axiom of Choice

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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 $F(\alpha)<F(\beta)$ holds...

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 $F(\beta)$.