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February 26th, 2018, 07:20 PM   #31
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 Originally Posted by topsquark Alright, let's try a bit of a detour. Perhaps it has value, perhaps it doesn't. We know that the rationals are countable. The reals consist of the rational numbers and the irrational numbers. The rationals are countable, and thus the irrationals must have the same cardinality as the reals because the irrationals can't be a finite set. (I hope we agree on that anyway.) So let's construct the table using only the set of irrationals... we no longer need to consider the rationals and thankfully the issue of 0.9999999... = 1 is no longer an issue. For convenience, let's use the section (0, 1) of real numbers. Then our list to "count" the irrationals is of the form: 0.00222459875978... 0.39457935730945... 0.79874968734978... etc. We are back to Cantor's diagonal argument as the simplest way to work with the list, which I know that zylo will just looooove. This table seems to me to be a cleaner argument than trying to take "limits" of real numbers. -Dan
Oh, so that's what this is about. I wonder if zylo's only qualm about the Cantor proof is that it uses decimal representations. Because I can easily give him a proof that doesn't use one. Maybe I should try this.

Last edited by skipjack; February 27th, 2018 at 12:49 AM. February 26th, 2018, 07:52 PM #32 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,660 Thanks: 2635 Math Focus: Mainly analysis and algebra It's not his only problem. As you can see, he doesn't understand how there can be infinitely many finite integers. I think it's because he won't let his preconceived ideas go and just follow what the logic says. February 26th, 2018, 09:35 PM   #33
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 Originally Posted by v8archie It's not his only problem. As you can see, he doesn't understand how there can be infinitely many finite integers. I think it's because he won't let his preconceived ideas go and just follow what the logic says.
I don't understand how you cannot understand that... But I don't really want to know the answer... February 27th, 2018, 01:09 AM   #34
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 Originally Posted by zylo I prefer to think of Sn as "Something" n, where "Something" is defined for every (all) n, for example, a proposition, definition, n-place decimal, number sequence,
In that statement, is "n" allowed to be infinite? If so, that should be stated at the outset, so as to avoid confusion. February 27th, 2018, 07:21 AM #35 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 n is never allowed to be infinite, it is only endless. But if that's what you mean, fine. I believe this is what's confusing some people: Let Sn =.999..9, n-places, by definition. $\displaystyle \lim_{n \rightarrow \infty}$Sn IS NOT, NOT 1, because of definition. F-lim = 1, for Function limit. $\displaystyle \lim_{n \rightarrow \infty}$Sn = .999999999999......, by definition. Call it D-lim for Definition limit. Then F-lim does not belong to {Sn} but D-limit does. That, by the way, is why you talk about the decimals in [0,1), not [0,1]. D-limit defines the real numbers. Last edited by zylo; February 27th, 2018 at 07:26 AM. February 27th, 2018, 08:41 AM   #36
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 Originally Posted by zylo n is never allowed to be infinite, it is only endless. Let Sn =.999..9, n-places, by definition. D-limit defines the real numbers.
In the above definition of Sn, is n finite? February 27th, 2018, 10:18 AM #37 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,660 Thanks: 2635 Math Focus: Mainly analysis and algebra Do you remember at school when you were about 6 years old and you were taught that 123 means one hundred, two tens and three units? Similarly $0.999\ldots9$ to $n$ places means $\sum \limits_{k=1}^n \frac9{10^k}$ and the infinite decimal $0.999\ldots$ means $$\sum \limits_{k=1}^\infty \frac9{10^k} = \lim_{n \to \infty} \sum \limits_{k=1}^n \frac9{10^k}$$. Your D-limit $S$ clearly isn't a member of $\{S_n\}$ because all of the $S_n$ terminate (after $n$ digits). $S$ does not. Thanks from topsquark Last edited by v8archie; February 27th, 2018 at 10:39 AM. February 27th, 2018, 02:36 PM   #38
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 Originally Posted by Micrm@ss So, the real numbers are often defined to be equivalence classes of rational Cauchy sequences. The hyperreal construction is very similar, but instead of taking Cauchy sequences, we take all sequences (incidentally, if we start with Q and do the hyperreal construction on it, then the result will contain R, it is an alternative construction to R!!) But the set of all sequences isn't a field, indeed, what would the inverse of (1,0,1,0,1,0,...) be? To solve this, we introduce the ultrafilter which decides for us which elements to identify and which not (this is problematic since ultrafilters cannot be constructed at all, and different choices of ultrafilters give different hyperreals!!)
Yes this is all coming back to me now.

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 Originally Posted by Micrm@ss A real sequence $(x_n)_n$ would then have as hyperreal limit the exact same sequence. I find this very elegant. And it gives some surprising other results: If we interpret 0.999999... as the limit of 0.9999...9, then this limit will be LESS than one, less by an infinitesimal amount.
Grrrrrr. I really hate this factoid, which is repeated in the Wiki article on .999... It's used by many people in the interminable .999... wars online, but it's misleading. As your notation indicates, there is a final 9. In fact your notation would be better written as $0.999 \dots 9_H$ where $H$ is some particular hyperinteger. The digit at all subsequent hyperinteger positions is zero. So this is more like saying (in the standard reals) that .999 < 1. That's true. And according to the Lightstone paper, using an argument that I followed at the time, that the expression consisting of ALL 9's does not represent a hyperreal at all.

This example is the basis for professor Katz's article, "A Strict Inequality .999... < 1," often referenced (by people who have read the title but not the article) as a valid argument that .999... = 1 is false. However if one reads professor Katz's article in detail, which I did at the time I got interested in all this, Katz does in fact admit that there is a final 9 at some hyperinteger index, after which all the remaining digits are zero. And that he is making more of a pedagogical point than a mathematical one.

My understanding is that by the transfer principle, .999... = 1 must be a theorem of the nonstandard reals just as it is in the standard reals. So this particular example you gave is misleading (although of course it's true) and only gives ammunition to the .999... cranks.

In fact I'm not sure I understand your remark about the "limit of the expression" being less than 1. Can you elaborate on that please? I am not sure that's correct. My understanding is that the limit of the expression is not a hyperreal at all. In fact this is an example of a nonempty set of hyperreals bounded above that lacks a least upper bound. The hyperreals are not topologically complete and this is an example of that problem to the best of my understanding.

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 Originally Posted by Micrm@ss The limit of the sequence (1,2,3,4,...) would be the hyperreal (1,2,3,4,...) and this hyperreal is different from (1, 12, 123, 1234,...) since their agreement set (there is only one place where it agrees, the first place) is finite, hence cannot lie in the ultrafilter.
Yes this all came back to me after I re-read a paper on NSA and gave this a little thought. A non-principal ultrafilter can not contain any finite sets, and must contain all cofinite sets. So in fact since we have $n < S_n$ for all $n > 1$, it must be the case that $[(n)]_U < [(S_n)]_U$ where $U$ is any nonprincipal ultrafilter and the square brackets indicate the equivalence class of the given sequence mod $U$.

All is clear now. This is a very enlightening example. Thanks for bringing it up.

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 Originally Posted by Micrm@ss Now, I have to be fair and say that this is not really how limits are defined in non-standard analysis. In non-standard analysis, we "fix" the issue that 0.999...=/=1 by basically saying that the limit operator should forget about the infinitesimal part. But to me the infinitesimal part is the nicest part, so I like my interpretation.
Yes but you agree that the row of 9's only goes out as far as some particular hyperinteger $H$, and that it's zero's after that ... right? So that this should NOT be taken as validation by the .999... cranks. That's my only point about this, that the hyperreals are frequently invoked (without understanding) by the .999... = 1 deniers, and it's very tedious to try to explain to them why the hyperreals don't help their case. Particularly in light of professor Katz's technically correct but misleading paper.

Last edited by Maschke; February 27th, 2018 at 02:47 PM. February 27th, 2018, 03:18 PM   #39
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 Originally Posted by Maschke My understanding is that by the transfer principle, .999... = 1 must be a theorem of the nonstandard reals just as it is in the standard reals. So this particular example you gave is misleading (although of course it's true) and only gives ammunition to the .999... cranks.
The transfer principle is the fudging of results by ignoring the infinitesimal part. It exists only to make the hyperreals useful as a tool of real analysis (which is good because that's why they were conceived). That's my understanding anyway.

Micrm@ss, being apparently my sort of purist, doesn't much like the fudge and prefers to consider analysis in the hyperreals divorced from real analysis. There may be a problem here as I understand that basic arithmetic breaks down in the hyperreals, but qualitative results are possible. More on this later.

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 Originally Posted by Maschke Yes but you agree that the row of 9's only goes out as far as some particular hyperinteger $H$, and that it's zero's after that ... right?
I'm far from convinced of that. For one thing, I'm unaware of any infinitesimal being quantified in any way that would make sense under a positional number system. Once you get past the integer indexed places, each "digit" could take any finite value. I'm prepared to believe that the coefficient of all higher powers of $\epsilon$ is zero (indeed, that probably ought to come out as part of the limit calculation). February 27th, 2018, 04:24 PM   #40
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 Originally Posted by v8archie I'm far from convinced of that. For one thing, I'm unaware of any infinitesimal being quantified in any way that would make sense under a positional number system.
Lightstone notation. https://en.wikipedia.org/wiki/A._H._Lightstone

The Wiki article illustrates my point that the 9's end at some particular hyperinteger position H. You'll also note that @Micrm@ss notated his number as 0.999...9. The lack of dots at the end is significant. Although I'm not sure if Micrm@ss was referring to the Lightstone notation. Lightstone notation has a semicolon in it so I'm not really sure what Micrm@ss intended here.

Quote:
 Originally Posted by v8archie Once you get past the integer indexed places, each "digit" could take any finite value.
Unclear what you mean by this.

Quote:
 Originally Posted by v8archie I'm prepared to believe that the coefficient of all higher powers of $\epsilon$ is zero (indeed, that probably ought to come out as part of the limit calculation).
What is $\epsilon$ in this context? It's certainly not the "smallest infinitesimal" or anything like that. The hyperreals are a field so there is no smallest positive hyperreal, standard or infinitesimal.

Last edited by Maschke; February 27th, 2018 at 04:29 PM. Tags limit, number Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post zylo Calculus 13 May 31st, 2017 12:53 PM Shen Elementary Math 2 June 5th, 2014 07:50 AM date Calculus 3 June 12th, 2012 11:51 AM kiv864 Applied Math 0 November 2nd, 2010 05:38 PM Anson Complex Analysis 1 February 16th, 2010 04:25 PM

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