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In the proof that Zorn's Lemma implies the Well-Ordering Theorem, it is required that the set $\mathcal{W}$ of all well-ordered subsets of some set $\mathcal{A}$ be partially ordered by continuation. (Then by showing that every chain in $\mathcal{W}$ has an upper bound, it follows from Zorn's Lemma that there exists a maximal element in $\mathcal{W}$, and it is shown that that element is a well-ordering of $\mathcal{A}$).

My question is, why is it necessary for $\mathcal{W}$ to be partially ordered by continuation, and not just by inclusion? Meaning $\left(C,\leq _C\right)\leq _{\mathcal{W}} \left(D,\leq _D\right) \iff C\subseteq D\;$ and $ \leq _C\; \subseteq\; \leq_D$ (without the additional requirement that $C$ be an initial segment of $D$).

I don't see why Zorn's Lemma can't apply in a similar way here too, and if it does why it doesn't follow that the maximal element in $\mathcal{W}$ is a well-ordering of $A$.

Thanks

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Call your poset $\mathcal{U}$ (since it's not quite $\mathcal{W}$). Zorn's lemma will definitely apply to $\mathcal{U}$ if we can show that chains have upper bounds. Clearly one natural upper bound of a chain $C$ - if such a thing exists - should be the union of its elements, so this amounts to:

Is the increasing union of a sequence of well-orderings guaranteed to be well-ordered?

It turns out that the answer is no (and so Zorn's lemma doesn't apply to your $\mathcal{U}$). The simplest example of this is probably the integers: although obviously $(\mathbb{Z}, <)$ isn't a well-ordering, each of the intervals $[-n, n]\cap\mathbb{Z}$ (for $n\in\mathbb{N}$) is well-ordered by $<$, and $\mathbb{Z}$ is the union of these well-orderings.

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If you just order by inclusion, it is no longer true that every chain has an upper bound. For instance, suppose $\mathcal{A}=\mathbb{Z}$. Then the sets $\{0\},\{-1,0\},\{-2,-1,0\},\dots$ are all well-ordered using the usual order and form a chain under inclusion, but their union is not well-ordered and so they have no upper bound in $\mathcal{W}$.

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  • $\begingroup$ Thanks a lot! For some reason, I assumed that a union of such a chain will always be well-ordered. $\endgroup$ – Evan Jan 26 '18 at 22:06

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