Timeline for Why are there no gaps in real numbers?
Current License: CC BY-SA 4.0
28 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 26, 2020 at 4:25 | vote | accept | Caleb Briggs | ||
| Oct 26, 2020 at 2:21 | comment | added | Mark S. | If by "chain" you mean any two numbers are comparable, then the rationals, the reals, and the surreals, all form a "chain". For how a chain could be uncountable (as in the reals), see math.stackexchange.com/q/1182145/26369 If by "chain" you mean any two numbers are comparable and you can, say, index them by integers and the order of the numbers in the chain agrees with the order of the integers, then not even the rationals form a chain in your sense. | |
| Oct 26, 2020 at 2:18 | answer | added | Noah Schweber | timeline score: 2 | |
| Oct 26, 2020 at 2:10 | history | edited | Caleb Briggs | CC BY-SA 4.0 | added 1192 characters in body |
| Oct 26, 2020 at 2:09 | answer | added | user326210 | timeline score: 3 | |
| Oct 26, 2020 at 2:05 | history | edited | Caleb Briggs | CC BY-SA 4.0 | added 1192 characters in body |
| Oct 26, 2020 at 2:02 | review | Close votes | |||
| Oct 26, 2020 at 18:38 | |||||
| Oct 26, 2020 at 1:57 | comment | added | Mark S. | What do you mean by "the surreals...are completely ordered...the real numbers can't be completely ordered"? This sounds backwards: the real numbers have an order that is Dedekind complete and the surreals don't. There is a least upper bound of $\{-1,-1/2,\ldots\}$ (negative reciprocals of naturals) in the reals (it's $0$), but there's no least upper bound in the surreals. | |
| Oct 26, 2020 at 1:55 | comment | added | Caleb Briggs | @MichaelBarz $\epsilon$ may not be real, but isn't it still a number? (at the very least, in the surreal numbers, it obeys all the rules of addition and multiplication and also can be expressed as a set). And isn't it true that there is supposed to be a real number that works as a bound for any set of rational numbers, or are there more restrictions? | |
| Oct 26, 2020 at 1:52 | comment | added | Ian | Well ordering says every subset has a least element, which also means that every subset of a well ordered set is well ordered. | |
| Oct 26, 2020 at 1:51 | comment | added | Caleb Briggs | @Ian Oh, so does well ordering require every subset also be well ordered, and so no intervals between real numbers can be well ordered? | |
| Oct 26, 2020 at 1:49 | comment | added | Ian | Another problem with your thinking is the assumption that there is a unique rational between two "neighboring" reals; actually between any pair of distinct reals there are already infinitely many rationals. | |
| Oct 26, 2020 at 1:48 | comment | added | Ian | $[a,\infty)$ and $(-\infty,a]$ are also not well ordered by the usual order on the reals. $[a,\infty)$ does have a least element, but its subset $(a,a+1)$ doesn't. | |
| Oct 26, 2020 at 1:46 | comment | added | Ng Chung Tak | I mean your meaning of "order", whether you can order a set by means of mapping an infinite set from natural numbers. As a result, there are equal amounts of odd numbers as natural numbers. | |
| Oct 26, 2020 at 1:46 | comment | added | user147556 | The problem is that $\epsilon$ isn't a real number. There is no real number that works as a bound. | |
| Oct 26, 2020 at 1:46 | comment | added | Noah Schweber | @CalebBriggs Just because there are a lot of numbers left at each stage, doesn't mean that we don't throw them all away eventually. Think about natural numbers: for each $n\in\mathbb{N}$ there are infinitely many $k>n$, but there are no natural numbers bigger than every $n\in\mathbb{N}$. | |
| Oct 26, 2020 at 1:45 | history | edited | Andrés E. Caicedo | edited tags | |
| Oct 26, 2020 at 1:44 | comment | added | Caleb Briggs | @NoahSchweber I seems like there are larger number that fit between 1/n and 0. All steeps along the way, there would exist some number $\epsilon$ stuch that 1/n > $epsilon$ > 0. So 0 seems like its too large to be the minimum lower bound, since it seems like there are larger numbers that would also work as a bound | |
| Oct 26, 2020 at 1:44 | comment | added | Calvin Khor | what do you mean by "completely ordered"? | |
| Oct 26, 2020 at 1:41 | comment | added | Caleb Briggs | @NgChungTak I edited the question. I mean to write a countably infinite number of steps not sets | |
| Oct 26, 2020 at 1:39 | history | edited | Caleb Briggs | CC BY-SA 4.0 | added 2 characters in body |
| Oct 26, 2020 at 1:38 | comment | added | Noah Schweber | "Consider {1,1/2,1/3,...}. The infimum seems as though it should be greater than 0." Why? | |
| Oct 26, 2020 at 1:37 | comment | added | Ng Chung Tak | Do you mean countable infinite set, cardinality, etc? | |
| Oct 26, 2020 at 1:35 | comment | added | Caleb Briggs | @littleO By "fully ordered" I mean put into one large chain of greater than signs. So something like r1 < r2 < r3 < r4 < .... < rn. Each object is ordered in a way that everything to the left of it is less, and everything to the right is greater | |
| Oct 26, 2020 at 1:31 | comment | added | littleO | What do you mean by "fully ordered"? | |
| Oct 26, 2020 at 1:28 | comment | added | Caleb Briggs | Would the reals still not be well ordered if we started at some fixed value. Like all reals greater than or equal to -10 for example? Or all non-negative reals? | |
| Oct 26, 2020 at 1:25 | comment | added | Ian | The reals are definitely totally ordered by their usual order. What they aren't is well ordered by that order. | |
| Oct 26, 2020 at 1:24 | history | asked | Caleb Briggs | CC BY-SA 4.0 |