Timeline for When a set of measure zero plus itself contains interior
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 22, 2012 at 16:03 | comment | added | Will Sawin | There is a natural map $A x A \to A + A$. If A is something such that "dimension" =is meaningful, such as a CW complex, then $A \times A$ has twice the dimension of $A$, and $A+A$ has no more than that dimension. Making $A$ curved will have no significant effect on the argument. Making $A$ totally disconnected will obviously make the connected components not have 3-dimensional convex hulls. I'm not sure which one you want. | |
| May 22, 2012 at 9:51 | comment | added | spr | The edges of polyhedron is 1 dimensional while the space is 3-dimensional. Can this difference of 2 dimensions be the reason? Precisely, is there a one-dimensional set $A$ for which $A+A$ can have interior in $\mathbb R^3$? Does the situation change if we substitute the edges of the polyhedron by something which does not contain an interval? | |
| May 22, 2012 at 8:49 | vote | accept | spr | ||
| Oct 13, 2012 at 9:46 | |||||
| May 21, 2012 at 12:22 | vote | accept | spr | ||
| May 21, 2012 at 12:22 | |||||
| May 21, 2012 at 10:56 | vote | accept | spr | ||
| May 21, 2012 at 10:56 | |||||
| May 21, 2012 at 9:17 | comment | added | Willie Wong | @Will: I was surprised at your answer to question #2 until I realised that we read the question differently. I parsed OP's "so does $A+A$" to mean "the convex hull of the connected component of $A+A$ also contains interior" which would of course be trivially true. | |
| May 21, 2012 at 6:47 | history | answered | Will Sawin | CC BY-SA 3.0 |