Timeline for Theorems that are 'obvious' but hard to prove
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 28, 2011 at 15:54 | comment | added | Deane Yang | Does this mean that it is hard to show that a rectangle does not have measure zero? | |
| Feb 28, 2011 at 15:52 | comment | added | Deane Yang | Yes, thanks, Mark, for the explanation. | |
| Feb 28, 2011 at 15:39 | comment | added | Jim Conant | Aha. That makes sense. | |
| Feb 28, 2011 at 15:34 | history | edited | Daniel Moskovich | CC BY-SA 2.5 |
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| Feb 28, 2011 at 3:34 | comment | added | Mark | I suppose that this is trivial if one constructs Lebesgue measure by first assigning measure to rectangles and in the end declares a set to be of measure zero if...well, if it has Lebesgue measure zero. If one merely has the classical definition of a zero measure set (a set such that $\forall \varepsilon > 0$ can be covered by countably many rectangles so that the sum of their volumes is at most $\varepsilon$), this is nontrivial. Actually Gowers already mentioned an equivalent result, in which the difficulty is more clear. | |
| Feb 28, 2011 at 2:50 | comment | added | Jim Conant | I'm also missing the point. I always thought this was a trivial proposition. | |
| Feb 28, 2011 at 2:14 | comment | added | Deane Yang | I must be completely missing the point here. Why is this hard? Can't you just show that there is a small cube inside $S$? | |
| Feb 21, 2011 at 19:43 | history | answered | Daniel Moskovich | CC BY-SA 2.5 |