# Proving that Zorn's lemma does not imply there exists well-ordered chains that contain their own upper bounds

We have the following statement:

Zorn's lemma is equivalent to the following statement: every partially ordered ser $$\langle A,<_R\rangle$$ has a chain which contains is own upper bounds

I have proved this first part quite easily, following an argument similar to the one that is used while proving the equivalence between Zorn's lemma and Hausdorff's maximal principle.

However, the corresponding exercise has a second statement:

If in the previous statement we add that the chain is well ordered, does the equivalence still hold?

I think that is not the case, but I don't know how to prove it.

It would be pretty interesting to have a partially ordered set that satisfies the Zorn's lemma hypothesis, but is such that every subset of it is not well-ordered with the restriction of the original partial order. That would make it, however, is such a thing possible? If not, how can we prove the equivalence does not hold? I am completely clueless at this point.

There are two key observations here:

1. A chain that contains its own upper bounds either contains a maximal element, or does not have a strict upper bound.

2. Zorn's lemma is equivalent to Zorn's lemma where we require that the chains are well-ordered.

The first observation is quite trivial. The second, however, lies within the standard proof of Zorn's lemma. Think about it.

How do we prove Zorn's lemma? We start with a partial order, and we construct a well-ordered chain by choosing, one step at a time, an upper bound. Then we argue, we had to have come to a stop, which provides us with a maximal element.

So to prove Zorn's lemma we only need to focus on well-ordered chains.

Your root mistake is, I think, misunderstanding the meaning of "well-ordered" here. When we say, for example, that $$A$$ is a well-ordered subset of $$\Bbb R$$, we usually mean by the standard order of the reals (which therefore implies $$A$$ is countable). So here when we say well-ordered chain in the partial order, we mean that the inherited order is indeed a well-ordering, and not just a linear ordering.

I mean, if you allow any well-ordering of the chain, then what is the point of having a partial order underlying the discussion to begin with? The question starts to lose meaning. And to some extent, is trivial now, since every well-ordered set, indeed any linearly ordered set, contains all of its upper bounds in itself.