Timeline for A density on the natural numbers invariant with respect to the multiplication
Current License: CC BY-SA 2.5
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 1, 2011 at 20:28 | vote | accept | Valerio Capraro | ||
| Mar 22, 2011 at 23:09 | comment | added | Gerry Myerson | @Gerald, isn't that already included in "finitely additive"? (and, thus, Kevin's example isn't an example?) | |
| Mar 22, 2011 at 19:30 | answer | added | Mark | timeline score: 7 | |
| Mar 22, 2011 at 16:55 | comment | added | Gerald Edgar | @Kevin: Or we could just require: if $\mu(A), \mu(B), \mu(A \cup B)$ all exist and $A \cap B = \emptyset$, then $\mu(A \cup B) = \mu(A) + \mu(B)$. | |
| Mar 22, 2011 at 16:29 | comment | added | Kevin O'Bryant | Another example: assign $\mu(A)=1$ if $A$ is infinite, and $\mu(A)=0$ if $A$ is finite. I suggest the following addendum to avoid trivialities: We should also insist that the following is a theorem: if $\mu(A)>0$, then $A$ contains arbitrarily long geometric progressions. | |
| Mar 22, 2011 at 5:29 | answer | added | Daniel Litt | timeline score: 4 | |
| Mar 22, 2011 at 5:05 | comment | added | Daniel Litt | @Gerry: True, and rereading the question it seems like the OP does not intend $0$ to be a natural number as well. | |
| Mar 22, 2011 at 4:31 | comment | added | Gerry Myerson | @Daniel, for many of us, $0$ is not a natural number. | |
| Mar 22, 2011 at 0:59 | comment | added | Daniel Litt | One example would be a measure assigning $\mu(A)=1$ if $A$ contains $0$, and $\mu(A)=0$ otherwise. | |
| Mar 21, 2011 at 23:50 | answer | added | Gerald Edgar | timeline score: 0 | |
| Mar 21, 2011 at 23:48 | comment | added | Gerald Edgar | That is, one can write down two sets that have density, but whose intersection does not have density. | |
| Mar 21, 2011 at 22:37 | comment | added | Did | Any finitely additive probability measure (also called a content) is defined on a field. But the collection of subsets $A$ of the natural numbers having a density $d(A)$ is not a field. | |
| Mar 21, 2011 at 21:52 | comment | added | Aaron Meyerowitz | when you say invariant with respect to the sum do you mean that $\lbrace k+2 \mid k \in A \rbrace$ has the same measure as $A$? | |
| Mar 21, 2011 at 20:09 | history | asked | Valerio Capraro | CC BY-SA 2.5 |