Work in ZF + DC. Assume that there is an uncountable $A \subseteq \mathbb{R}$ and a linear order $\prec$ on $A$ such that for every $x \in A$, $\{y \in A: y \prec x\}$ is countable. Must there exist an injection from $\omega_1$ to $\mathbb{R}$? If true, perhaps one should try to construct a $\prec$-cofinal subset $B \subseteq A$ such that $(B, \prec)$ is well-ordered. But I do not see how to do this with dependent choice.
1 Answer
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You can't do this.
Working over the Solovay model, take a symmetric extension as follows. First, force by adding $\Bbb Q\times\omega_1$ many Cohen reals, with order automorphisms (lexicographic order, that is) acting on the forcing with the filter of groups generated by countable supports. So the set of reals is still linearly ordered.
This preserves $\sf DC$, and we can show that any function from the an ordinal into the reals must be given by adding a single Cohen real.
So, the set of reals we added is linearly orderable, every proper initial segment is countable, but there is no injection from $\omega_1$ into the reals.
answered May 27 at 22:20
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$\begingroup$ What about the case where such an order exists on $\mathbb R$ itself? $\endgroup$ Commented May 27 at 23:29
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$\begingroup$ Oh, that's an interesting question. I don't know. $\endgroup$– Asaf Karagila ♦Commented May 29 at 14:17
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1$\begingroup$ @GabeGoldberg Assume $V=L,$ let $\kappa$ be inaccessible, let $G=\langle G_{\alpha} : \alpha < \kappa \rangle$ be Levy collapse generic. Let $R=\mathbb{R}^{L[G]}$ and define $f: R \rightarrow \kappa$ by setting $f(x)$ to be the least $\alpha$ such that $x \in L[G \restriction \alpha].$ Then $M=L(R, f)$ has $(R, \prec)$ as in the question, and every set of ordinals in $M$ is OD from $f$ and some $G \restriction \alpha,$ but $R \cap HOD^M(f, G \restriction \alpha) = R \cap L[G \restriction \alpha]$ is countable in $M.$ So no subset of $\omega_1$ codes an injection from $\omega_1$ to $R.$ $\endgroup$ Commented Aug 28 at 4:42