How do you show, assuming the Axiom of Choice and the Continuum Hypothesis, that there exists a well-ordering on $[0,1]$ such that for all $x$, there are only countably many $y$ such that $y \leq x$?
Existence of a particular well-ordering of [0,1]
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0Good; then you can probably make sense of Arturo's answer, though you may need to give it a bit of thought. – 2012-04-30
2 Answers
If CH holds and AC both hold, then $[0,1]$ (which is bijectable with $\mathbb{R}$, hence with $2^{\aleph_0}$) is bijectable with $\omega_1$, the first uncountable ordinal. Let $f\colon [0,1]\to\omega_1$ be a bijection, and define the order with $x\leq y\iff f(x)\preceq f(y)$ (the right hand side is the usual ordering of ordinals).
Since $\omega_1$ is the first uncountable ordinal, every element of $\omega_1$ has only countably many elements strictly smaller than it, so for every $\alpha\in\omega_1$, $\{a\in\omega_1\mid a\preceq\alpha\}$ is countable. Thus, for any $x\in [0,1]$, only countably many reals can be strictly smaller than $x$ in this ordering.
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0Ah, ok. Thanks! – 2012-04-30
Following Munkers 10.2. Set $S(a)=\{ x | x. If $W= \emptyset$ we are done. Otherwise W has a least element $\Omega$. Now $S( \Omega )$ is as required.
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0$S(\Omega)$ is an uncountable subset of [0,1] by the CH it is in bijection with [0,1]. The answer to the second question is yes. – 2014-04-28