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Aug 24, 2023 at 20:58 review Close votes
Aug 31, 2023 at 3:09
Aug 24, 2023 at 20:49 comment added Ross Ure Anderson I did not receive any responses on MSE, however I did receive a very helpful answer on here, which also provides a non-measure theory solution to a similar problem cited in this answer. And yes, compactness is used. By 'interval' I mean any kind of interval in $\mathbb{R}$, including unbounded.
Aug 24, 2023 at 20:43 comment added Nate Eldredge One thing to consider is that your proof has to use the completeness of $\mathbb{R}$ in some form, since the statement is not true for $\mathbb{Q}$ (the unit interval in $\mathbb{Q}$ can be covered by a sequence of intervals whose total lengths are less than 1, or indeed arbitrarily small). The most common proofs use it via the fact that the unit interval in $\mathbb{R}$ is compact. But it means the proof has to be somehow "topological" and can't be done with pure arithmetic.
Aug 24, 2023 at 20:39 comment added Nate Eldredge First question: when you say "interval", do you mean open, closed, half-open, what?
Aug 24, 2023 at 20:38 comment added Nate Eldredge I think Math.SE is a better place for this question, since it's standard textbook material.
Aug 24, 2023 at 19:42 answer added Fedor Petrov timeline score: 2
Aug 24, 2023 at 17:05 history edited Ross Ure Anderson CC BY-SA 4.0
added 343 characters in body
Aug 17, 2023 at 15:10 answer added Ross Ure Anderson timeline score: 1
Aug 17, 2023 at 12:46 comment added ho boon suan The answer at mathoverflow.net/a/51555 and its comments seem relevant to Problem 1.
Aug 16, 2023 at 23:50 history edited YCor CC BY-SA 4.0
removed capitals from title
Aug 16, 2023 at 22:28 answer added Nik Weaver timeline score: 5
Aug 16, 2023 at 21:47 history asked Ross Ure Anderson CC BY-SA 4.0