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Let $(L,<)$ be a linear order, and $S\subseteq L$ be bounded below.

Question. Is it true that there is a countable set $\{s_i\mid i\in \mathbb N\}\subseteq S$ such that for all $x\in S$, there is some $s_i\leq x$?

Thoughts:

  • If the answer is no, what assumptions are needed for this to hold?
  • Let $L=\mathbb{N}^{\mathbb{N}}/\asymp$, with the equivalence relation $\asymp\,=\{(f,g):f=\Theta(g)\}$. For two classes $[f],[g]\in L$, we put $[f]<[g]$ if $f=o(g)$, and complete this to a linear order. Does $(L,<)$ have the aforementioned property? Edit: this post suggests not.
Logan Post
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1 Answers1

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No it's not true. Let $L = \omega_1+1$ have the order $x < y$ iff $y\in x$ and let $S = \omega_1$. Then $S$ is bounded from below by $\omega_1$, but if $s_i\in S$ then $s = \bigcup_i s_i + 1$ is such that $s < s_i$ for all $i$ yet $s\in S$.

The property you are asking for is that $S$, as a linear order, has countable coinitiality. Note that $S$ being bounded from below in $L$ is actually irrelevant, since you can always add a bottom element to any linear order. So all there is to do is find a set with uncountable coinitiality (or dually, cofinality). The dual order on $\omega_1$ is such example.

Jakobian
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