Open set in $\mathbb{R}$ as countable union of open intervals and which version of Choice

Asked Jan 30 '15 at 23:37

Active Jan 30 '15 at 23:38

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In proving, every non-empty open set in $\mathbb{R}$ is union of a countable collection of disjoint open intervals in $\mathbb{R}$.

It seems to me this result is using some version of Choice(probably AC). Am I correct is it using some version of AC(maybe weaker) or not?

edited Jan 30 '15 at 23:38

Daniel W. Farlow

asked Jan 30 '15 at 23:37

Sushil

I think there might be at least one other duplicate. But that one came up first. *Edit:* Found another one. – Asaf Karagila Jan 30 '15 at 23:42
(The short answer, no, you don't use the axiom of choice, because the rational numbers can be well-ordered without using the axiom of choice.) – Asaf Karagila Jan 30 '15 at 23:43

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