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I was reading the following How do i prove that every open set in $\mathbb{R}^2$ is a union of at most countable open rectangles?

The answer by "user4594" basically considers a k-cell with rational endpoints, and I am guessing that when he says that "E would be the union of these cells" the axiom of choice is needed because for each x there may be more than one k-cell with rational endpoints.

Am I correct?

Community
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mononono
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2 Answers2

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You could actually take all rectangles with rational endpoints containing $x$, since there are at most countably many of them, so you don't need choice.

Anyway, choice on subsets of a countable set is provable without the full Axion of Choice.

Thomas Andrews
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  • But, isn't the "index set", which in this case the set itself, uncountable? – mononono Mar 15 '15 at 01:38
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    No, you can well-order the set of all rectangles with rational coordinates. Then, for any $x$, you can take the least rectangle in that well-ordering that contains $x$, which is well-defined.In any event, you don't need to choose one, you can take all of the rational rectangles for each $x$. @Dejon – Thomas Andrews Mar 15 '15 at 01:41
  • (I might have used the wrong term when I wrote countable choice - the Axiom of Countable Choice is not what I was referring to...) – Thomas Andrews Mar 15 '15 at 01:44
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    Basically, $U$ is the union of all rational rectangles contained within it, and the set of those rectangles is countable. – Thomas Andrews Mar 15 '15 at 01:57
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    Ah, well, at least I'm in good company with the random comment-less downvoter also hitting Asaf. Sigh – Thomas Andrews Mar 15 '15 at 02:06
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    Good company is hard to find nowadays! – Asaf Karagila Mar 15 '15 at 02:09
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Without the axiom of choice the rational numbers can be well-ordered, since they are countable. Therefore the set of $k$-rectangles with rational endpoints is countable.

So we don't need the axiom of choice to really choose from this collection.

Asaf Karagila
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