# Understanding the upper bound in Zorn's Lemma

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 $$\Sigma$$?

• 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 $[0,1]\ne\cup C.$) Jan 16 at 22:32

Yes. You are absolutely right. We need to prove that the union of the chain is in $$\Sigma$$. Sometimes it's not.
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. $$[0,1]\cap\Bbb Q$$.