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Subgroup of $\mathbb{R}$ either dense or has a least positive element?

If I have $G$ a closed subgroup of $\mathbb{R}$, then why is $G$ necessarily countable, except of course in the case where $G=\mathbb{R}$?

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Addison
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    Direct consequence of the result at http://math.stackexchange.com/questions/90177/subgroup-of-mathbbr-either-dense-or-has-a-least-positive-element – Jonas Meyer Dec 10 '11 at 23:19
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    Addison: If you can see how to apply the result in the linked question to solve your problem, you could post a solution to your own question here. – Jonas Meyer Dec 10 '11 at 23:36
  • Thank you Jonas, I will think about it for a bit. – Addison Dec 10 '11 at 23:39
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    Well, I don't quite see it. Would appreciate an answer, only of course if someone finds the time. – Addison Dec 11 '11 at 00:14
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    It can be dangerous to use "any" as a synonym of "every". "If any member of the club so requests, the treasurer will make the records available" changes its meaning if you say "every". "Prove that any real number is purple" can mean "Pick any real number and prove that it's purple" (you might do that if some real numbers aren't purple). I think Paul Halmos may have urged people never to use the word "any" in this kind of context. – Michael Hardy Dec 11 '11 at 00:49
  • I've tried to fix the wording. – Addison Dec 11 '11 at 00:55
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    If the subgroup is dense and closed then must be everything. If it's not dense, it has a least positive element and the arguments in the linked thread show that that this element generates the subgroup. – t.b. Dec 11 '11 at 01:13

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