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In the linked question here, the user demonstrates two examples of extension of morphisms using Zorn's Lemma arguments, and I've seen the same pattern to extend morphisms before in other sources.

However, none of them contain a verification of the condition of Zorn's lemma that every totally ordered subset has an upper bound. They jump straight from establishing the order relation to the existence of a maximal element.

Presumably this is because the verification follows a routine pattern that is obvious if you have seen it before. But supposing I haven't, how do we know that these posets meet the upper bound condition of Zorn's Lemma?

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  • $\begingroup$ In many cases it's considered "obvious" so the author doesn't give the details. . For example, let $F$ be the set of free (non-principal) filters on an infinite set $S$. Let $F$ be partially ordered by $f<g\iff f \subsetneqq g$. If $C$ is a non-empty $<$-chain then it is easily shown that $\cup C\in F$ and $\cup C$ is a $<$-upper-bound for $C.$ ....So $F$ contains a $<$-maximal member, i.e. a free ultra-filter on $S$ exists.... Note that I didn't mention another necessary but obvious point: That $F$ is not empty. $\endgroup$ – DanielWainfleet Dec 31 '17 at 17:10
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You are correct that the verification is a necessary part of applying Zorn's lemma and it is often skipped because it follows a routine pattern.

The setup is generally that we have a chain $\{(A_i, f_i)\}_{i \in I}$ where the $A_i$ are sets and the $f_i\colon A_i \to X$ are morphisms satisfying $A_i \subseteq A_j$ and $f_i = f_j|_{A_i}$ whenever $(A_i, f_i) \leq (A_j, f_j)$. Then let $A = \bigcup_iA_i$ and define $f\colon A \to X$ by the following rule: For $a \in A$ choose $i$ such that $a \in A_i$ and define $f(a) = f_i(a)$.

Now you have to check two things, first our definition of $f$ required a choice and we need to prove $f$ does not depend on this choice. This follows because of the restriction conditions on the $f_i$ which imply that $f_i(a)$ is the same for all $i$ such that $a \in A_i$. Second, you have to check that $(A, f)$ is an upper bound for the chain $(A_i, f_i)$. This amounts to showing that $f|_{A_i} = A_i$, which is true by definition of $f$ (because we can choose that $i$).

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  • $\begingroup$ Very clear, thank you! $\endgroup$ – Sam Cassidy Dec 30 '17 at 15:32
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    $\begingroup$ Since you are showcasing a technical detail of how Zorn's lemma applies to the inclusion-extension order, you might want not to use the sloppy notation $A_0\subseteq A_1\subseteq\cdots$ to indicate a chain. I think you should stay true to a chain being a totally ordered subset of the set you want to prove inductive, without making the assumption of such a chain being well-ordered (or even isomorphic to $\Bbb N$) as well. $\endgroup$ – user228113 Dec 30 '17 at 15:41
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    $\begingroup$ And instead of talking about a choice of $i$, just define $f=\bigcup_i f_i$, and argue that it's a function with whatever properties you need. $\endgroup$ – Henning Makholm Dec 30 '17 at 15:56
  • $\begingroup$ Changed the notation to accomodate non-linear chains. I will definitely not change it to use the notation $\bigcup_if_i$ though. Appologies to set theorists but in most areas of mathematics we don't think of functions as being equal to their graphs. Proving that $\bigcup_if_i$ is a function is equivalent to proving that $f_i(a)$ is independent of the choice of $i$, so the union notation does not simplify the argument at all, it just places the argument in a context that is less familiar to most mathematicians. $\endgroup$ – Jim Dec 31 '17 at 14:53

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