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Ollie
Ollie
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Turing Turning linear ordering into well-ordering

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Ollie
Ollie
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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: x \prec y\}$$\{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.

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 $x \in A$, $\{y \in A: x \prec y\}$ 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.

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.

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Ollie
Ollie
  • 33
  • 3

Turing linear ordering into well-ordering

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 $x \in A$, $\{y \in A: x \prec y\}$ 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.