Timeline for Subsets of $\mathbb{R}^+$ closed under addition
Current License: CC BY-SA 4.0
19 events
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| Oct 9, 2022 at 17:01 | history | edited | LSpice | CC BY-SA 4.0 |
Capitalise title, and name of linked question, while this is on the front page
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| Oct 9, 2022 at 11:59 | answer | added | Salvo Tringali | timeline score: 2 | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Feb 20, 2014 at 8:49 | history | edited | user9072 |
edited tags
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| Apr 3, 2012 at 10:25 | comment | added | Emil Jeřábek | ... each of which must also lie in $[0,nc]$. Since a continuous image of a compact set is compact, the set of sums of $m$ elements of $[0,nc]\cap C$ is compact (and therefore closed) for each $m\le n$, hence $[0,nc]\cap C^+$ is closed. It follows that $C^+$ itself is closed. | |
| Apr 3, 2012 at 10:21 | comment | added | Emil Jeřábek | @Benjamin: You have now answered your first question (the only sets not covered by the generic case of Anton's constructions are those which fail to be closed after removal of $0$, i.e., those that have $0$ as a limit point. You've show the only one such is $[0,\infty)$.) For the second question, if $C$ is a closed set of positive reals, its closure under addition $C^+$ is automatically topologically closed. To see this, let $c=\min(C)>0$, which exists as long as $C\ne\emptyset$ because $C$ is closed. Let $n>0$. Any point in $[0,nc]\cap C^+$ must be the sum of at most $n$ elements of $C$, ... | |
| Apr 3, 2012 at 0:45 | comment | added | Benjamin Steinberg | The point is if $x_n$ is a sequence converging to 0 and (a,b) is an interval with a>0 then we can find $x_m<b-a$. Now choose a positive integer k with $kx_m>a$ with k smallest possible. The $kx_m$ is in that interval. | |
| Apr 3, 2012 at 0:35 | comment | added | Benjamin Steinberg | If a closed semigroup contains 0 as a limit point then it is all positive reals. I will put up a reference later. | |
| Apr 3, 2012 at 0:21 | comment | added | Benjamin Steinberg | If you take the closed subsemigroup generated by the elements $1+1/\sqrt{p}$ with p prime has 1 as the only rational number in it less than 2. I don't yet see how to do something like this near 0. | |
| Apr 2, 2012 at 23:25 | comment | added | Misha | OK, that's clear too, no proper perfect closed submonoids. | |
| Apr 2, 2012 at 23:06 | comment | added | Misha | The only mildly interesting question here is if ${\mathbb R}_+$ contains a closed proper sub-semigroup where $0$ is not an isolated point. In particular, is there a totally-disconnected perfect subset of ${\mathbb R}_+$ which is a sub-monoid? | |
| Apr 2, 2012 at 22:58 | comment | added | Benjamin Steinberg | What does he mean by the only set remaining? Also does one not have to close again under the topology? | |
| Apr 2, 2012 at 22:53 | comment | added | Emil Jeřábek | Anton’s comment gives a recipe how to construct all closed subsets of $\mathbb R^+$ closed under addition. | |
| Apr 2, 2012 at 22:44 | comment | added | Benjamin Steinberg | I don't understand either comment. Help? | |
| Apr 2, 2012 at 22:42 | comment | added | Pietro Majer | In particular, if that closed set $C$ has a minimum positive element, then it generates a closed additive semigroup $S:= C\cup (C+C) \cup (C+C+C)\dots$. If $C$ has no minimum positive element, it generates a dense semigroup. | |
| Apr 2, 2012 at 22:39 | answer | added | user6976 | timeline score: 4 | |
| Apr 2, 2012 at 22:31 | comment | added | Anton Petrunin | You can take any closed set of positive reals, close it with respect to addition. After that you can include $0$ if you want to. The only remaining set will be $[0,\infty)$. | |
| Apr 2, 2012 at 22:29 | answer | added | Benjamin Steinberg | timeline score: 2 | |
| Apr 2, 2012 at 22:09 | history | asked | Michael Hardy | CC BY-SA 3.0 |