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Related to my answer on this question : set of values of finite measure, exhaustion method

I would like to know if for an arbitrary total order $K$, there exists a sequence $\{k_n\}$ such that for all elements $k \in K$, there exists $k_n$ such that $k \le k_m$. In some way, it would allow you to "go to infinity".

The reason why I needed this is because I was trying to use Zorn's Lemma in order to answer a question on MSE, and I came up with this set $$ \bigcup_{k \in K} E_k $$ where $E_{k_1} \subseteq E_{k_2}$ whenever $k_1 \le k_2$ and $K$ is a total order. The $E_k$'s were elements of a $\sigma$-field, so if I have the sequence $k_n$, I can re-write this union as a countable union and show that the above set is in the $\sigma$-field. (That is what I needed.)

Any suggestions?

Community
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Patrick Da Silva
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2 Answers2

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If by sequence you mean something indexed by the natural numbers, and $K$ is truly arbitrary, the answer is no.

For example, let $K$ be the countable ordinals, with the natural order. There is no countable set cofinal with $K$.

For something nearer to what you are interested in, let $c$ be the cardinality of the continuum. Produce a well-ordering of $\mathbb{R}$ by finding a one to one correspondence between $\mathbb{R}$ and the ordinals $\lt c$. Starting with the empty set, add the reals, one at a time, in the order defined by the well-ordering of $\mathbb{R}$. We get a set $K$ of subsets of $\mathbb{R}$, totally ordered by the inclusion relation, with no countable cofinal subset.

André Nicolas
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  • Exactly the counter example I wanted. Thanks – Patrick Da Silva Aug 06 '11 at 14:47
  • What are the elements of $K$? Your argument confuses me. I am not acquainted with ordinals. Mostly because the real numbers do have a countable cofinal subset, so I don't see what you're doing... – Patrick Da Silva Aug 06 '11 at 15:00
  • @Patrick Da Silva: The real numbers do have a countable cofinal subset, if we use the natural ordering. I used a highly unnatural ordering, the one obtained from well-ordering $\mathbb{R}$, using the ordinals less than $c$. I cannot define the ordinals and derive their basic properties in this limited space. I suspect there is a half-decent summary on Wikipedia. I know there is a good one in Halmos' Naive Set Theory. – André Nicolas Aug 06 '11 at 15:08
  • Oh. Anyway. It wasn't for my own purposes, so I guess I'll just trust you on this one and read about it later if I want to. =) Thanks – Patrick Da Silva Aug 06 '11 at 15:33
  • @Patrick Da Silva: In case the title of Halmos' classic book puts you off, he meant it somewhat tongue in cheek, as contrasted with Axiomatic Set Theory. He could have called it Set Theory for all Mathematicians. – André Nicolas Aug 06 '11 at 22:19
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As André explains in his answer, your questions asks whether every linear order has countable cofinality. Let's consider a particular type of linear orders, namely cardinals. A cardinal is regular if its cofinality is equal to itself. André gave an example of an uncountable regular cardinal, namely $\omega_1$. The latter can be defined (depending on your axioms) as the minimal uncountable cardinal (the minimal one exists since the ordinals are well-ordered, and by Cantor's theorem $2^\omega$ is uncountable, so there are some uncountable ordinals).

Like any other ordinal, it consists of all ordinals smaller than it, all of which are countable. Because of the former property, a subset of $\omega_1$ (or any other ordinal) is cofinal if its union is $\omega_1$. Cantor's diagonal argument shows that the union of countably many countable sets is countable, hence $\omega_1$ has uncountable cofinality (and by minimality, it is regular).

Conversely, a cardinal like $\aleph_\omega$ is singular (not regular) since it is the union of $\aleph_0,\aleph_1,\aleph_2,\ldots$.

Unrelated, one may ask whether there is a scale for your partial order [thanks, ccc, for the correction]. This is a linearly-ordered cofinal set. We can assume (exercise) that any scale is well-ordered. Given a partial order $\alpha$, define its bounding number $\mathfrak{b}(\alpha)$ as the minimal cardinality of an unbounded set in $\alpha$, and its dominating number $\mathfrak{d}(\alpha)$ as the minimal cardinality of a cofinal set in $\alpha$. It's not hard to see (exercise) that $\mathfrak{b}(\alpha) \leq \mathfrak{d}(\alpha)$. Moreover, equality holds if there is a scale for $\alpha$ (but the converse is not necessarily true, see Asaf's comment).

ZFC is not enough to determine whether some simple and natural orders have scales, for example various orders on sequences. Usually, assuming CH an easy construction shows that there is a scale, whereas under Cohen's classical model there is none. The subject studying these properties is called cardinal characteristics of the continuum. For a highlight, check out Cichoń's diagram.

Yuval Filmus
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  • It might be worth pointing out, though, that in this context of linear orders everything in the second half of your answer degenerates. Every linear order has a scale (itself!), sets are unbounded iff they are cofinal, etc. These characteristics start to become interesting in the more general context of partial orders. (Also, as an aside, in my experience the phrase "diagonal argument" usually refers to the trick showing that the reals are uncountable rather than the claim (using some choice!) that countable unions of countable sets are countable.) – user83827 Aug 06 '11 at 16:39
  • To slightly add a bit on ccc's comment, it is worth noting that the axiom of choice is not needed to show the existence of uncountable ordinals. For more information on that see my answer here. – Asaf Karagila Aug 06 '11 at 16:48
  • Consider the partial order on $\mathbb N\times{0,1}$ defined as $(m,n)<(k,l)$ if and only if $m<K$ in the usual sense, and $n=l$. In this order there is no cofinal set which is linearly ordered. I believe you need some directness in order to have a scale, either that or you need to add the definition of a scale. :-) – Asaf Karagila Aug 06 '11 at 22:20
  • @Asaf: Your example can be generalized by picking two infinite cardinals $\kappa < \lambda$ and taking for the second coordinate all $\kappa$-subsets of $\lambda$; now the condition on the second coordinate is $\subseteq$. This is $\kappa$-directed and $\mathfrak{b} = \mathfrak{d} = \lambda$ but there is no scale. – Yuval Filmus Aug 07 '11 at 19:44
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    @Yuval: Nice! I was looking for the simplest avail, though, to make my point simple. :-) – Asaf Karagila Aug 07 '11 at 19:46