Timeline for On the independence of lower and upper asymptotic and Banach densities
Current License: CC BY-SA 3.0
6 events
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| May 17, 2015 at 10:10 | comment | added | Gabriel Dill | I've edited my answer to include @MartinSleziak's comment and fix a mistake (the first case requires $\beta > \alpha$, not $\beta > 0$, in order to work; this required adjusting the following cases). I know that this answer is probably not very convincing; unfortunately I haven't been able to write down the general construction without getting bogged down in too many technical details. Anyway, I hope it does at least somewhat illuminate the problem; if it's only confusing, I will gladly delete it. | |
| May 17, 2015 at 10:01 | history | edited | Gabriel Dill | CC BY-SA 3.0 |
fixed a mistake
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| May 17, 2015 at 9:41 | comment | added | Salvo Tringali | @Gabriel. I've at least two issues with your answer. First, as you too have remarked, it doesn't answer my question. Second, I'm not so convinced that, after filling in the technical details that you're alluding to, the construction you are suggesting will still be so fairly easy: I might mention a number of situations where constructions involving densities are relatively easy (or even trivial to some degree) as long as the "relevant parameters" are rational, but get significantly more complicated otherwise. In any case, thanks for your contribution to the discussion. | |
| May 17, 2015 at 8:55 | history | edited | Gabriel Dill | CC BY-SA 3.0 |
fixed typo
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| May 17, 2015 at 8:50 | comment | added | Martin Sleziak | Changing a set $A$ to $B=\{\lfloor ca \rfloor; cA\}$ for $c\in(0,1)$ changes all four densities with factor $c$. So it is sufficient to prove the case $\delta=1$. (Which is not much of a simplification, but perhaps it helps at least a bit.) | |
| May 17, 2015 at 8:47 | history | answered | Gabriel Dill | CC BY-SA 3.0 |