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March 6th, 2018, 11:29 AM   #31
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Quote:
 Originally Posted by cjem The issue is that the usual definition of a limit makes use of real numbers (at least implicitly). So defining real numbers in terms of these limits is completely circular.
I thought it was the OP who said that. The reals are defined first, and then limits are defined afterward. That invalidates OP's point.

March 6th, 2018, 11:43 AM   #32
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Quote:
 Originally Posted by cjem The issue is that the usual definition of a limit makes use of real numbers (at least implicitly). So defining real numbers in terms of these limits is completely circular.
No. The definition of limit works perfectly well with rationals. You only need real numbers for the limit itself. I believe that most places I see the definition written, no domain is specified (for $\epsilon$, which is the key quantity here).

March 6th, 2018, 11:53 AM   #33
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Quote:
 Originally Posted by v8archie You only need real numbers for the limit itself.
Exactly. Not sure why you're bringing up the domain of $\epsilon$ - that was never in question. The point is that the definition of a limit makes use of real numbers insofar as that the limits themselves are real numbers... In any case, a definition of real numbers referring to limits in this way is circular.

Last edited by cjem; March 6th, 2018 at 12:01 PM.

March 6th, 2018, 12:10 PM   #34
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 Originally Posted by v8archie No. The definition of limit works perfectly well with rationals. You only need real numbers for the limit itself. I believe that most places I see the definition written, no domain is specified (for $\epsilon$, which is the key quantity here).
Brings up an interesting subtlety.

Referring back to the Wiki page on the Cauchy construction of the rationals, they define a rational Cauchy sequence as:

"... for every rational ε > 0 ..."

Aha! The Wiki author has thought this through correctly.

To define the reals as Dedekind cuts, we don't need any epsilons, only sets of rationals. No problem. Having defined the reals that way, we can define Cauchy sequences using arbitrary epsilons. The domain of the epsilons is never important.

But if we want to define the reals as equivalence classes of Cauchy sequences of rationals, we have to first carefully define Cauchy sequences using epsilons taken over the rationals.

Then you can define the reals. And having done that, you can define limits of real sequences with epsilon being real. In the end none of this matters because the rationals are dense in the reals. So you get the same definition of limits whether epsilon is rational or real.

But to be picky, if we are using the Cauchy sequence construction of the reals, we have to start by specifying that epsilon is rational. And Wiki got this detail right.

Last edited by Maschke; March 6th, 2018 at 12:27 PM.

 March 6th, 2018, 01:52 PM #35 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 Real Number Definition: The infinite sequence of digits $\displaystyle \lim_{n \rightarrow \infty}.a_{1}a_{2}...a_{n}$ is a real number in [0,1). Two real numbers are the same iff each digit is the same. The algebraic limit of a real number exists but the real number is not unique. Unique real numbers: digital $\displaystyle \lim_{n \rightarrow \infty}$.49999.... digital $\displaystyle \lim_{n \rightarrow \infty}$.50000.... algebraic $\displaystyle \lim_{n \rightarrow \infty}$.49999....=1/2 BUT algebraic $\displaystyle \lim_{n \rightarrow \infty}$.49999....9m=1/2 for all m. Definition: algebraic $\displaystyle \lim_{n \rightarrow \infty}\frac{a_{1}}{10^{1}}+\frac{a_{2}}{10^{2}}+. ....+\frac{a_{n}}{10^{n}}$ Test: Are the real numbers .4999999.... and .50000000...... the same? Are their algebraic limits the same? It's really quite simple.
March 6th, 2018, 04:31 PM   #36
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Quote:
 Originally Posted by zylo Definition: The infinite sequence of digits $\displaystyle \lim_{n \rightarrow \infty}.a_{1}a_{2}...a_{n}$ is a real number in [0,1).
This use of the limit notation is not defined until you clearly define it. And without a clear definition of the real numbers, I don't know what a real number in [0,1) is. And if 9 is a digit, why isn't your interval [0,1]?

Also the phrase, "algebraic limit" is quite contradictory in this context. Analysis differs from algebra in that analysis involves limits and algebra doesn't. Now (to satisfy the pedants, of which I'm one) it's true that there are "limits" in category theory such as inverse limits and colimits, but these are not relevant to the current context.

And ... I know you've been asked this before, but ... since you'd read some Rudin, and since you've tried to study the real numbers, why haven't you grokked anything?

Last edited by Maschke; March 6th, 2018 at 04:36 PM.

March 6th, 2018, 05:23 PM   #37
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Quote:
 Originally Posted by zylo Definition: The infinite sequence of digits $\displaystyle \lim_{n \rightarrow \infty}.a_{1}a_{2}...a_{n}$ is a real number in [0,1).
In saying that you are assuming that 0.99999... is less than 1 which is what everyone here has been denying.

Quote:
 Two real numbers are the same iff each digit is the same.
No, that is not true. 1.0000... = 0.9999..., exactly what is being argued here. If you want to use it to show that 1.0000... is NOT 0.9999... then you will have to prove this statement, not just assert it.

Quote:
 The algebraic limit of a real number exists but the real number is not unique.
I don't know what that means. I know what the "limit of a sequence of numbers" means but I don't know what you could mean by the "limit" of a single number. And the limit of a sequence of real numbers is unique.

Quote:
 Unique real numbers: digital $\displaystyle \lim_{n \rightarrow \infty}$.49999.... digital $\displaystyle \lim_{n \rightarrow \infty}$.50000....
Yes, each of those limits is a real number- in fact they are the same number, 1/2.

Quote:
 algebraic $\displaystyle \lim_{n \rightarrow \infty}$.49999....=1/2 BUT algebraic $\displaystyle \lim_{n \rightarrow \infty}$.49999....9m=1/2 for all m.
I'm not clear what the "m" represents here. Is that sequence .49m, .499m, ..4999m, etc.?

Are you trying to make a distinction between "digital limit" and "algebraic limit"? If so what is that distinction.

Quote:
 Definition: algebraic $\displaystyle \lim_{n \rightarrow \infty}\frac{a_{1}}{10^{1}}+\frac{a_{2}}{10^{2}}+. ....+\frac{a_{n}}{10^{n}}$
That's not a "definition". A definition would say what that is equal to or what it is called.

Quote:
 Test: Are the real numbers .4999999.... and .50000000...... the same? Are their algebraic limits the same? It's really quite simple.
Yes, it is. They are both equal to 1/2. The first can be written as 0.4+ 9(.01)+ 9(.001)+ 9(.0001)+ ...= 0.4+ 0.09(1+ .1+ .01+ ...)= 0.4+ 0.09(1+ (1/10)^{-1}+ (1/10)^{2}+ ...+ (1/10)^n+ ....

A series of the form a+ ab+ ab^3+ ...+ ab^n+ ... is called a "geometric series". It is fairly easy to show that such a series has sum [math]\frac{a}{1- b} as long as |b| is less than 1.

Here, 1+ (1/10)+ (1/10)^2+ ...+ (1/10)^n+ .. has a= 1 and b= 1/10. The sum is 1/(1- 1/10)= 1/(9/10)= 10/9. .09(10/9)= .1 so the entire sum is 0.4+ 0.4= 0.5= 1/2. I hope I don't have to prove that .50000... is also equa; to 1/2!

March 7th, 2018, 06:24 AM   #38
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Quote:
 Originally Posted by zylo Unique real numbers: digital $\displaystyle \lim_{n \rightarrow \infty}$.49999.... digital $\displaystyle \lim_{n \rightarrow \infty}$.50000.... algebraic $\displaystyle \lim_{n \rightarrow \infty}$.49999....=1/2 BUT algebraic $\displaystyle \lim_{n \rightarrow \infty}$.49999....9m=1/2 for all m. Definition: algebraic $\displaystyle \lim_{n \rightarrow \infty}\frac{a_{1}}{10^{1}}+\frac{a_{2}}{10^{2}}+. ....+\frac{a_{n}}{10^{n}}$
I clearly defined what i meant by algebraic limit. I consider summation an algebraic operation. The point was that the real number is the unique infinite digital sequence, not the calculus limit of its algebraic sum.

The point was to distinguish between "limits," the failure to do so being the source of all the confusion.

March 7th, 2018, 06:49 AM   #39
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Quote:
 Originally Posted by zylo The point was that the real number is the unique infinite digital sequence, not the calculus limit of its algebraic sum.
But that is just wrong. You've just made it up.

 March 10th, 2018, 05:41 PM #40 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,515 Thanks: 2515 Math Focus: Mainly analysis and algebra On the subject of irrational numbers not existing in the real world: Formula for $\pi$ discovered in hydrogen atoms Wikipedia article on the Wallis product $$\frac21 \frac23 \frac43 \frac45 \frac65 \frac67 \cdots = \frac\pi2$$ Thanks from topsquark

 Tags 0.999, 0.999..., 0.999...=1, arguments, fail, finitism, real numbers, usual

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