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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 $a \in A, a\geq s$?

Sune Jakobsen
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    No. This is a homework problem. – Sam Nead Nov 26 '09 at 16:36
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    I agree, this is a homework problem: -1. – Alicia Garcia-Raboso Nov 26 '09 at 17:06
  • You should require $a>s$ (strictly greater than) for some $s$, otherwise you could have S bounded and $a$ a maximum. – Jose Brox Nov 26 '09 at 17:24
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    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 E Speyer Nov 26 '09 at 17:47
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    David - it is presented as a straight problem. Nothing about "I need this for..." Or "I was reading X and thought...". No motivation... However, I will be happy to be corrected by M. Jakobsen. – Sam Nead Nov 26 '09 at 17:54
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    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
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    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 E Speyer Nov 27 '09 at 00:16
  • Well, I apologize in all directions, in that case.

    Sune: would you like to explain the resolution of your refined intersection problem?

    – Sam Nead Nov 27 '09 at 00:40
  • I haven't solved it. But if the answer to my question was yes, you could find a countable subset $J\subset I$, such that for every $i\in I$ there is a $j\in J: K_j\subset K_i$. If there is a j that works for every i, the intersection would be $K_j$ and thus non-empty. Otherwise you could find a decresing sequence $K_1\supset K_2\supset \dots$ and reduce the problem to the homework problem. – Sune Jakobsen Nov 27 '09 at 07:04
  • I think that your generalization still holds. Perhaps transfinite induction will be useful? – Sam Nead Nov 27 '09 at 16:21
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    After Sune's explanation I am happy to withdraw my downvote. – Alicia Garcia-Raboso Nov 27 '09 at 19:57
  • Is the OP missing a claim about the size of the set? Isn't the empty set a counterexample for the first sentence? – thejoshwolfe Jun 07 '16 at 20:20

2 Answers2

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

http://en.wikipedia.org/wiki/Long_line_(topology)

Kristal Cantwell
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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
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    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 01 '09 at 17:05
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Keyword: cofinality.

http://en.wikipedia.org/wiki/Cofinality

Added MUCH later: To be slightly more explicit, for any cardinal $\kappa$, the cofinality of the successor cardinal $\kappa^+$ is $\kappa^+$, so not only can we not in general find an unbounded (=cofinal) countable subset, there is no fixed cardinality $\kappa$ such that every totally ordered set has an unbounded subset of cardinality at most $\kappa$.

Pete L. Clark
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  • care to explain? – thejoshwolfe Jun 07 '16 at 20:20
  • @thejoshwolfe: The OP's question is equivalent to: "does every totally ordered set have countable cofinality?" So learning about cofinality -- say from the linked to article -- will answer the question. – Pete L. Clark Jun 07 '16 at 21:46
  • i think you're supposed to actually answer the question in a stackexchange answer to a question. http://meta.stackexchange.com/a/8259/206371 cofinality seems like it would be a cool concept to use to answer the question, but I can't figure out how it would work. For example, I had no idea the OP's question was equivalent to your sentence; I wouldn't have figured that out on my own without spending hours reading several interlinking wikipedia articles. – thejoshwolfe Jun 08 '16 at 02:38
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    @thejoshwolfe: This answer is almost seven years old and in fact predates mathoverflow being part of the stackexchange system and predates the site math.SE. As it is an undergraduate level set theory question, if it were asked today it would probably be closed and/or migrated. The existence of totally ordered sets -- e.g. cardinals -- of uncountable cofinality is the answer to the question, and that is discussed in the article. If you have further questions, please feel free to ask on math.SE. – Pete L. Clark Jun 08 '16 at 03:17