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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0I think that is not possible. Given any $x=0,a_1a_2a_3a_4\dots a_n$, any $y=0,a_1a_2a_3a_4\dots a_na_j$ will do the job. Am I wrong? – 2012-04-30
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4Perhaps giving it the order type of $\omega_1$? – 2012-04-30
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0Could you explain further? – 2012-04-30
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0Are you familiar with infinite ordinal numbers? If you are, the construction is very easy; if not, it will take a bit more explanation. – 2012-04-30
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0I'm familiar with the 'basic' infinite ordinals like $\omega$, less so with $\omega_1$. – 2012-04-30
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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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0So I can see how you need CH to show that $\omega_1$ can be bijected with $\mathbb{R}$ (otherwise you just know that $\omega_1$ is uncountable), but I'm not sure how AC comes into play. – 2012-04-30
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2@Patrick: without some choice, you might not have a well ordering of $[0,1]$, so you couldn't have the bijection with $\omega_1$. – 2012-04-30
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2@Patrick: You need AC to be sure that $[0,1]$ ($\mathbb{R}$) is bijectable with a cardinal. In some models of ZF without Choice, $\mathbb{R}$ is a countable union of countable sets, and so cannot be bijected with a cardinal (since such a cardinal would necessarily be countable, but we know that $\mathbb{R}$ is not countable). – 2012-04-30
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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