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The starting point of this question is the observation that in ${\sf (ZFC)}$, all ordinals $\alpha < \omega_1$ can be order-embedded in $\mathbb{R}$.

Let $\omega^\omega$ denote the set of all functions $f:\omega\to\omega$. Given $f, g:\omega\to\omega$ we say $f\leq^* g$ iff there is $N\in\omega$ such that $f(k) \leq g(k)$ for all $k\geq N$.

We say ${\cal D}\subseteq \omega^\omega$ is dominating if for all $f\in \omega^\omega$ there is $d\in {\cal D}$ such that $f\leq^* d$, and we say ${\cal B}\subseteq \omega^\omega$ is unbounded if for all $f\in\omega^\omega$ there is $b\in {\cal B}$ such that $b\not\leq^*f$. (A diagonalization argument shows that every unbounded (and therefore also every dominating) family must be uncountable, and there are interesting set-theoretical considerations in this context, see here.)

We define ${\frak b}$ to be the smallest cardinality that an unbounded family ${\cal B}\subseteq\omega^\omega$ can have, and let ${\frak d}$ be the smallest cardinality that a dominating family ${\cal D}\subseteq\omega^\omega$ can have. It is consistent that $\omega_1 < {\frak b} < {\frak d} \leq 2^{\aleph_0}$.

Question. Is it provable in ${\sf (ZFC)}$ that ${\frak b}$ or even ${\frak d}$ can be order-embedded in $\mathbb{R}$? (The question could also be asked of other cardinal characteristics of the continuum.)

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    $\begingroup$ Isn't every ordinal embeddable in $\Bbb Q$ countable? What am I missing? $\endgroup$
    – Alessandro Codenotti
    Commented May 16, 2023 at 16:18
  • $\begingroup$ Right - will modify, I should have written \mathbb{R} $\endgroup$
    – Dominic van der Zypen
    Commented May 16, 2023 at 16:19
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    $\begingroup$ The same is true for $\Bbb R$ $\endgroup$
    – Alessandro Codenotti
    Commented May 16, 2023 at 16:22
  • $\begingroup$ If an infinite limit ordinal embeds into $\mathbf{R}$, its convex hull is order-isomorphic to $\mathbf{R}_{\ge 0}$, so changing the embedding we can suppose that its convex hull is $\mathbf{R}_{\ge 0}$. Now assuming the ordinal is $\omega_1$, we obtain an unbounded subset of $\mathbf{R}_{\ge 0}$ in which every countable subset is bounded. This is of course absurd. $\endgroup$
    – YCor
    Commented May 16, 2023 at 16:40
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    $\begingroup$ @YCor Note that as written that approach actually requires choice (it's consistent with ZF that there is an unbounded subset of the reals all of whose countable subsets are finite). It's better to just pick the lex-least rational between successive elements in the range of a putative embedding. $\endgroup$
    – Noah Schweber
    Commented May 16, 2023 at 17:10

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Assume that $i : \alpha \to \mathbb{R}$ is an order-embedding of some ordinal $\alpha$ into $(\mathbb{R},<)$. We can modify $i$ to yield an order embedding $j : \alpha \to \mathbb{Q}$ by induction over $\beta < \alpha:

The interval $[i(\beta), i(\beta + 1)]$ is a non-degenerate interval in $\mathbb{R}$, which contains no points from the range of $i$ other than its endpoints. Pick rationals $p,q$ with $i(\beta) \leq p < q \leq i(\beta + 1)$, and set $j(\beta) = p$ (if $\beta$ is a limit ordinal, otherwise $j(\beta)$ is already defined) and $j(\beta + 1) = q$.

Thus, the ordinals that order-embed into $\mathbb{R}$ are precisely the countable ones.

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    $\begingroup$ Or, just use the fact that there's a rational between $i(\beta)$ and $i(\beta+1)$ for each $\beta+1<\alpha$ to get a contradiction directly without porting over to $\mathbb{Q}$. $\endgroup$
    – Noah Schweber
    Commented May 16, 2023 at 17:05

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