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(Edit: The first formulation is wrong. See the second answer) Does every totally ordered set contain an unbounded countable subset. In other words: If S is a totally ordered set, can we find a (edit: at most) countable subset A, such that for every $s \in S$, there is a $a \in A, a\geq s$?

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No. This is a homework problem. – Sam Nead Nov 26 '09 at 16:36
I agree, this is a homework problem: -1. – Alberto García-Raboso Nov 26 '09 at 17:06
I am baffled as to why people think this is a homework problem. It could be assigned in a set theory class, but it is a very natural question and the counter-examples are not elementary. I'll bow to peer pressure and not give an explicit construction, but the basic hint here is to read up on ordinal numbers. – David Speyer Nov 26 '09 at 17:47
It was not a homework problem, but it was inspired by a homework problem. The problem was: Show that if $K_1\supset K_2\supset \dots$ is a decreasing sequence of non-empty compact sets, the intersection $\cap_{i=1}^{\infty} K_i$ is non-empty. I was wondering if this could be generalized: Let $(K_i)_{i\in I}$ be a system of non-empty compact sets, such that for for all $i,j\in I: K_i\subset K_j \vee K_i\supset K_j$. Is the intersection $\cap_{i\in I} K_i$ non-empty? This is why thought of the problem, but I got interested in the problem for its own rights. – Sune Jakobsen Nov 26 '09 at 18:26
For the record, I remember inventing and thinking about this question when I was first learning set theory. My motivation went as follows: An equivalent formulation of Zorn's lemma is "In a nonempty poset where every totally ordered subset has an upper bound, there is a maximal element." At the time I found it hard to think about arbitrary totally ordered sets, so I wondered if I could replace this by "In a nonempty poset where every ascending sequence $(a_i)_{i \in Z}$ has an upper bound, there is a maximal element." – David Speyer Nov 27 '09 at 0:16
up vote 4 down vote accepted

There is a counterexample in the long line L. It is totally ordered and every sequence has a limit in L. see the following:

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As long as we are answering homework questions (grump!) a simpler counter-example is provided by the first uncountable ordinal. – Sam Nead Nov 26 '09 at 17:46
The fact that the long line is a counterexample is completely derivative on the fact that omega_1 (the first uncountable ordinal) is a counterexample. Thus, Sam Nead's answer in the comment here is far better. – Joel David Hamkins Dec 1 '09 at 17:05

Keyword: cofinality.

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