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This is an exercise question from Chapter 2 of A Course in Galois Theory by D.J.H. Garling:

Suppose that $(A,\leq)$ is an infinite well-ordered set. Show that there is a unique element $a$ such that $\{x:x<a\}$ is infinite, while $\{x:x<b\}$ is finite for each $b<a$.

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Am I missing something here? $\mathbb{N}$ is well-ordered, but does not satisfy the described property...?

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If you’ve quoted the exercise exactly, you’re not missing a thing: $\Bbb N$ with the usual order is a counterexample. The exercise needs another hypothesis: there is at least one $a\in A$ such that $\{x\in A:x<a\}$ is infinite. – Brian M. Scott Dec 9 '12 at 12:00
In my copy of the book we have the additional assumption "... with a greatest element". That excludes $\mathbb N$ and makes the exercise solvable. Hint: If $a_0 = \max A$, can $\{x\mid x < a_0\}$ be finite? Now let $B = \{a \in A \mid \{x\mid x < a\} \text{ is infinite}\}$. – martini Dec 9 '12 at 12:06
I removed the axiom-of-choice tag. With well-orders given, AC is not needed (nor is Zorn's lemma, of course) – Hagen von Eitzen Dec 9 '12 at 12:15
@hwhm: If you don't believe the answers you get here, and you're in personal communication with the author, why not ask him instead? – Henning Makholm Dec 9 '12 at 15:36
What does Zorn's lemma have to do with this question, and why is it in the title? – Asaf Karagila Dec 9 '12 at 18:17
up vote 4 down vote accepted

Indeed there seem to have a typo there. In fact two typos.

The set $\mathbb N$, or $\omega$, is a counterexample to the first point. However if there is at least a single point $a\in A$ such that $\{b\mid b<a\}$ is infinite, then the conclusion follows. You can show that the order of $\{b\mid b<a\}$ is in fact isomorphic to $\omega$, and so it is not only a counterexample it is the only counterexample.

Similarly the part about an uncountable set also contains a mistake because $\omega_1$, the least uncountable ordinal, is defined to be the set of all countable ordinals. It can be shown that this set is uncountable, and every point has only countably many predecessors, but there is no point which has uncountably many predecessors. As before, this is the only counterexample to this statement.

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@hwhm: Yes. If every point has only countably many predecessors then there no point which has uncountably many predecessors; the other direction is as easy. – Asaf Karagila Dec 19 '12 at 5:08

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