20 Nov 14
Originally posted by twhiteheadtwhitehead in what way did you define integers as being a complete set? I don't remember ever learning that.
Please tell us more about this 'completeness'. The completeness I learned in school had the integers being a complete set.
What follows are a few shameless plugs of texts that I've written in the past and can tell you a little bit more about the meaning of the real numbers being a complete set.
The motivation of the completeness axiom is given in this post: http://climbingthemountain.wordpress.com/2008/12/06/real-analysis-basics-iii/ but I'd advise unfamiliar readers (I'm looking at you GreatKIngRat) to look at the whole series in order to make sense of a way in which we can construct the real numbers:
http://climbingthemountain.wordpress.com/2008/10/25/real-analysis-basics/
http://climbingthemountain.wordpress.com/2008/11/22/real-analysis-basics-ii/
http://climbingthemountain.wordpress.com/2008/12/06/real-analysis-basics-iii/
http://climbingthemountain.wordpress.com/2008/12/21/real-analysis-inductive-sets/
http://climbingthemountain.wordpress.com/2009/01/02/real-analysis-exercises/
These blog posts are somewhat formal and even a little bit dry but I think that if you make an honest effort you'll get to see why so many of us are enthralled with mathematics.
Then I also have written two more popular blog posts that try to explain the motivation of building up different systems of numbers while starting with the natural numbers and the sum operation. My main argument is that this motivation comes from trying to define new mathematical operations and defining sets of numbers that always satisfy these new mathematical operations.
So for instance we start with natural numbers and addition and then construct subtraction but while trying to do something simple like 5-7 we see that the result isn't defined in the natural numbers so that we have to "create" a new set of numbers in which subtraction always make sense. Etc
http://physicsfromthebottomup.blogspot.com/2009/07/lets-talk-about-numbers.html
http://physicsfromthebottomup.blogspot.com/2010/12/more-talk-about-numbers.html
Unfortunately there was a planned third post in this series that I have never got to write due to laziness and/or writer's block.
PS: I know you didn't ask me about it, but I'm just trying to shed some more light in this issue while shamelessly plugging my texts of yore.
20 Nov 14
Guys and Girls, I have very much enjoyed this thread, but I’m backing out of the discussion now. You have posted a lot of wonderful things and regardless of whether or not I accept that 0.999... = 1, it is all worth reading, which I will (and might get back to, in a different thread in a different time). I feel I should tell you that I’m stopping now, because I don’t want you to spend a lot of time writing stuff aimed at me without me reading it. Naturally, this thread can still continue on.
Note that I will read what you’ve written so far, although it may take some time.
So, Adam, Duchess, twhitehead (why can’t those two just get along??), DeepThought, Soothfast, thanks for your considerable efforts.
And please do carry on the thread of course.
20 Nov 14
Originally posted by adam warlockhttp://math.kennesaw.edu/~plaval/math4381/lubconsequences.pdf
twhitehead in what way did you define integers as being a complete set? I don't remember ever learning that.
Definition 177 A subset S of R is said to be complete if it satisfies the supremum
(and therefore the infimum) property. In other words S is said to be complete if every non-empty bounded subset of S has both a supremum and an infimum in S.
20 Nov 14
Originally posted by adam warlockBy definition or by a proof?
Ok, the set of integers is trivially complete.
I can come up with a number between 1 and 2: 1.5
(But this number is not an integer so that one doesn't count!)
Oh, I see.
Then I can come up with a number between any two nonequal rational numbers a and b arbitrarily close: (a+b)/2. This is a rational number, right? Then the rational set are complete, right?
(But there are reals between these two a and b as well, whichever you chose.)
Oh, but they are not rational so they don't count.
Okej, somewhere I am wrong. Where?
(I ask this question because I want to learn, not to imply that you are wrong.)
20 Nov 14
1 edit
Originally posted by FabianFnas
By definition or by a proof?
I can come up with a number between 1 and 2: 1.5
(But this number is not an integer so that one doesn't count!)
Oh, I see.
Then I can come up with a number between any two nonequal rational numbers a and b arbitrarily close: (a+b)/2. This is a rational number, right? Then the rational set are complete, right?
(But there ...[text shortened]... wrong. Where?
(I ask this question because I want to learn, not to imply that you are wrong.)
By definition or by a proof?
The definition of completeness allows one to prove trivially that the set of integers is trivially complete.
The document linked by twhitehead offers an argument that allows you to see that.
As for the set of rational numbers they are not complete in the sense defined above.
Take for instance the (canonical) set {x<sqrt(2)}. This set is clearly non-empty, it has an upper bound, but its supremum (sqrt(2)) is not a rational number. Hence the set of rational numbers isn't complete.
What your argument proves is that the set of rational numbers is dense: given any two rational numbers there always is a rational number between them (this definition is equivalent to the more technical definition of a dense set).
In conclusion you can have sets that are dense, but not complete; and you can have sets that are complete but are not dense.
PS: a little bit of googling found this neat explanation of why the integer numbers are complete but the rational numbers are not: http://www.thestudentroom.co.uk/showthread.php?t=1664297
20 Nov 14
Originally posted by adam warlockI can read in the link that "Q is not complete with respect to the Euclidean metric since there exist Cauchy sequences in Q which don't converge in Q." and it doesn't say anything for an amateur.By definition or by a proof?
The definition of completeness allows one to prove trivially that the set of integers is trivially complete.
The document linked by twhitehead offers an argument that allows you to see that.
As for the set of rational numbers they are not complete in the sense defined above.
Take for instance the (canoni ...[text shortened]... plete but the rational numbers are not: http://www.thestudentroom.co.uk/showthread.php?t=1664297
If you cannot explain it in a laymen terms, then please just say so.
I gave a question, and you don't answer them.
20 Nov 14
1 edit
Originally posted by FabianFnasI'll give the questions you've made in your post and my answer to them (most of them were given in my original post, but I'll rephrase them a little bit)
I can read in the link that "Q is not complete with respect to the Euclidean metric since there exist Cauchy sequences in Q which don't converge in Q." and it doesn't say anything for an amateur.
If you cannot explain it in a laymen terms, then please just say so.
I gave a question, and you don't answer them.
By definition or by a proof?
The definition of completeness allows one to prove trivially that the set of integers is trivially complete.
Hence it is by proof.
This is a rational number, right?
I didn't answer your question in my original post, but the answer is yes. It is a rational number.
Then the rational set are complete, right?
That I've answered in my original post. And the answer is no.
Okej, somewhere I am wrong. Where?
Yes you are. You are mixing up dense with complete. A set is said to be dense if between any of its two members you can still find a third element that is still a member of the set. A set is said to be complete if for all sequences that can be defined for members of the set their limits is always inside the set (I'm speaking rather informally).
Thus the set of rational numbers is dense and you already gave an algorithm that shows it. For it to be complete you'd have to prove that all sequences of rational numbers converge to a rational number. If you want to disprove that the rational numbers have one sequence that doesn't converge to a rational number.
Take "x" to be a rational number and define the set {x<\sqrt(2)}. With this set you can define a sequence of rational numbers that converges to an irrational number. Hence the rational numbers don't form a complete set.
20 Nov 14
Originally posted by twhiteheadYes, thinking about it they are complete. There is no Cauchy sequence which does not converge, because there are no Cauchy sequences. Every subset has a least upper bound because there are no open subsets (every non-empty subset contains it's limit points). I didn't check that statement when I was writing the post. It's a pretty poor form of completeness though.
That's what I thought. I just wanted to make sure as DeepThought seemed to be saying otherwise, and my memory is a little shaky - its been over 20 years since I did that in school.
20 Nov 14
Originally posted by adam warlock
I'll give the questions you've made in your post and my answer to them (most of them were given in my original post, but I'll rephrase them a little bit)
By definition or by a proof?
[b]The definition of completeness allows one to prove trivially that the set of integers is trivially complete.
Hence it is by proof.
[quote]Th ...[text shortened]... rs that converges to an irrational number. Hence the rational numbers don't form a complete set.[/b]
You are mixing up dense with complete.That was the mistake I made in the post that sparked this.
20 Nov 14
Originally posted by DeepThoughtActually I was doing it too until I really to stop to think about the definition of completeness (the link provided by twhitehead is what really made it click!).You are mixing up dense with complete.That was the mistake I made in the post that sparked this.
But after that click it is pretty obvious that the set of integers is trivially complete. If I ever realized this in my days of being a student it is safe to say that I had forgot about it completely until now.
20 Nov 14
Originally posted by adam warlock"A set is said to be complete if for all sequences that can be defined for members of the set their limits is always inside the set"
I'll give the questions you've made in your post and my answer to them (most of them were given in my original post, but I'll rephrase them a little bit)
By definition or by a proof?
[b]The definition of completeness allows one to prove trivially that the set of integers is trivially complete.
Hence it is by proof.
[quote]Th ...[text shortened]... rs that converges to an irrational number. Hence the rational numbers don't form a complete set.[/b]
There is beauty here. I know it. But I see no beauty in your explanation.
Sorry I asked. I just wanted to know. To see the beauty.