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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 $y \leq x$?

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  • $\begingroup$ I 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? $\endgroup$
    – Pedro
    Apr 30, 2012 at 17:28
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    $\begingroup$ Perhaps giving it the order type of $\omega_1$? $\endgroup$ Apr 30, 2012 at 17:29
  • $\begingroup$ Could you explain further? $\endgroup$
    – Venge
    Apr 30, 2012 at 17:31
  • $\begingroup$ Are you familiar with infinite ordinal numbers? If you are, the construction is very easy; if not, it will take a bit more explanation. $\endgroup$ Apr 30, 2012 at 17:32
  • $\begingroup$ I'm familiar with the 'basic' infinite ordinals like $\omega$, less so with $\omega_1$. $\endgroup$
    – Venge
    Apr 30, 2012 at 17:34

2 Answers 2

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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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  • $\begingroup$ So 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. $\endgroup$
    – Venge
    Apr 30, 2012 at 17:40
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    $\begingroup$ @Patrick: without some choice, you might not have a well ordering of $[0,1]$, so you couldn't have the bijection with $\omega_1$. $\endgroup$ Apr 30, 2012 at 18:00
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    $\begingroup$ @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). $\endgroup$ Apr 30, 2012 at 18:08
  • $\begingroup$ Ah, ok. Thanks! $\endgroup$
    – Venge
    Apr 30, 2012 at 18:25
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Following Munkers 10.2. Set $S(a)=\{ x | x<a \}; W =\{x | S(x)\text{ uncountable}\}$. If $W= \emptyset$ we are done. Otherwise W has a least element $\Omega$. Now $S( \Omega )$ is as required.

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  • $\begingroup$ $S(\Omega)$ 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. $\endgroup$
    – user139964
    Apr 28, 2014 at 7:08

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