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I am trying to prove that Hausdorff Max. Principle can be proved if we assume Zorn's lemma.

I have found the next proof on the internet:

Let $X$ be a partially ordered set and let $C = Ch(X)$, ordered by $⊆$, where $Ch(X)$ is just the set of all chains in $X$.

Let $D ∈ Ch(X)$ and $E = ∪D$.

Then $E ∈ Ch(X)$.

$E$ is an upper bound for $D$ so, by Zorn’s Lemma, $C$ has a maximal element.

What I don't get is why $E \in Ch(X)$, wouldn't that imply that any partial order is actually inductive partial order, which is quite nonsensical to claim.

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1 Answer 1

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The proof you have in your question is wrong. It should be $D\subseteq Ch(X)$, and not $D\in Ch(X)$. We care about a chain of chains, rather than a chain in $X$.

The fact that the union of a chain of chains is again a chain can be checked easily: if $x,y\in E$, then there is some $C\in D$ such that $x,y\in C$ and since $C$ is a chain, they are comparable.

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