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Assume $ \Sigma $ is a partially ordered set with the $\subseteq $ relation.

Assume $C \subseteq \Sigma$ is a chain. Then we can deduce that $ b = \bigcup_{a\in C}^{}a$, is an upper bound for the chain $C$. And this would be true for any chain in $\Sigma$.

I'm assuming I have some conceptual misunderstanding. Because from what I know we should prove that the upper bound $b$ is a member of $\Sigma$. But since the upper bound $b$ equals $ \bigcup_{a\in C}^{}a$, doesn't it mean that the upper bound is a member of the chain which is a subset of $\Sigma$. Meaning that $b\in \Sigma$? Why should I have to prove that the upper bound b is a member of $\Sigma$?

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    $\begingroup$ We do not know (a priori) that $\cup C\in\Sigma$. E.g. let $\Sigma$ be the set of all subsets of $[0,1]$ that have largest members. Then $C=\{[0, 1-1/n]:n\in \Bbb N\}$ is a chain but $\cup C\not \in \Sigma $ (... $[0,1]$ is the only upper bound in $\Sigma$ for $C$ but $[0,1]\ne\cup C.$) $\endgroup$ Jan 16 at 22:32

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Yes. You are absolutely right. We need to prove that the union of the chain is in $\Sigma$. Sometimes it's not.

In the vast majority of cases, however, at least the "trivial" cases, this is not a problem. For example, the increasing union of linearly independent sets in some fixed vector space, is itself a linearly independent set.

But indeed, this is why we argue that this union is linearly independent, by showing that given any finitely many vector in said union, they appear somewhere in the chain, and therefore must be linearly independent.

We can easily concoct examples where the union of the chain is not necessarily in the model. Just take $\Sigma$ to be all the initial segments of $[0,1]\cap\Bbb Q$ of the form $[0,q]\cap\Bbb Q$ for some $q\in\Bbb Q$, including the full set itself, for example. Then look at all the initial segments whose members are all smaller than $1/e$, or any other irrational number. The union of this chain is an initial segment of all the rational numbers below $1/e$, which is not in $\Sigma$. But the chain still has an upper bound, e.g. $[0,1]\cap\Bbb Q$.

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  • $\begingroup$ This is fantastic, thank you very much. $\endgroup$
    – Nizar
    Jan 16 at 20:42

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