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Yes. This is an easy exercise: For every $r\in\Bbb R$ fix some sequence of rational numbers $r_n$ such that $\lim r_n=r$. Now enumerate $\Bbb Q$ as $\{q_n\mid n\in\Bbb N\}$ and consider $A_r=\{k\mid\ …

You seem to flip some directions of the order, take $\Bbb N$, every chain has a lower bound, but there is no set of maximal elements. You also want to quantify over all $X$, in the sense that choice i …

Assuming the axiom of choice, the answer is yes. Pick an element $x_0\in P$. If $x_0$ is not maximal then $\lbrace x\in P\mid x_0 < x\rbrace$ is nonempty. We can choose some $x_1$. Suppose for $\alp …

Note that two words of the same lengths are comparable if and only if they are equal. So you can order the words in the following way: $$X\prec^* Y\iff |X|<|Y|\text{ or } (X<_{\rm Lex}Y \text{ and } | …

Yes. This was essentially proved by Honsel and Forti in the 1980s by analysing a model that generalises the Cohen model (essentially, the one Monro used to show it can be consistent for Dedekind finit …

There are $2^{\aleph_0}$ subsets of $\Bbb Q$ which are order isomorphic to $\Bbb Q$. To see this, note that $\Bbb{Q\setminus N}$ is order isomorphic to $\Bbb Q$, and consider for every $A\subseteq\Bb …