February 26th, 2018, 01:43 PM  #21  
Senior Member Joined: Oct 2009 Posts: 263 Thanks: 90  Quote:
S1 = 1, ending S2 = 2, ending S3 = 3, ending S100 = 100, ending. For which n is Sn nonending? Last edited by skipjack; February 27th, 2018 at 12:40 AM.  
February 26th, 2018, 01:59 PM  #22 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,282 Thanks: 93 
0r, let Pn be the number of nplace decimals in [0,1) listed as .0000, .0001, ........, .9999 = 9999. P= $\displaystyle \lim_{n \rightarrow \infty}$Pn, a member of {Pn} ={9, 99, 999, 9999, ..............} What is the largest member of {Pn}? I am not asking what any particular member of {Pn} is, I am asking what the largest member is. Last edited by skipjack; February 27th, 2018 at 12:40 AM. 
February 26th, 2018, 01:59 PM  #23  
Senior Member Joined: Oct 2009 Posts: 263 Thanks: 90  Quote:
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!!) 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. 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. Now, I have to be fair and say that this is not really how limits are defined in nonstandard analysis. In nonstandard 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. Last edited by skipjack; February 27th, 2018 at 12:43 AM.  
February 26th, 2018, 02:01 PM  #24  
Senior Member Joined: Oct 2009 Posts: 263 Thanks: 90  Quote:
 
February 26th, 2018, 02:16 PM  #25  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,282 Thanks: 93  Quote:
How did you manage to post your long post the instant afterI posted mine? Coincidence I guess. "rubbish thread?" Can't do any better than that? Last edited by zylo; February 26th, 2018 at 02:28 PM.  
February 26th, 2018, 02:53 PM  #26 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,686 Thanks: 666 Math Focus: Wibbly wobbly timeywimey stuff. 
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 Last edited by skipjack; February 27th, 2018 at 12:50 AM. 
February 26th, 2018, 03:45 PM  #27 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,214 Thanks: 2410 Math Focus: Mainly analysis and algebra  
February 26th, 2018, 03:55 PM  #28 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,214 Thanks: 2410 Math Focus: Mainly analysis and algebra  I don't think this guarantees that your diagonal constructs an irrational. Indeed, it seems easy to guarantee that it doesn't: make every element $0.00\ldots012345678910111213\ldots$ with $n$ zeros at the start of the $n$th element.

February 26th, 2018, 05:05 PM  #29  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,214 Thanks: 2410 Math Focus: Mainly analysis and algebra  Quote:
Clearly, there isn't one. All $t_n \gt t_{n1}$ and all $t_n \lt 2$. So $t \not \in \{t_n\}$.  
February 26th, 2018, 07:14 PM  #30  
Senior Member Joined: Oct 2009 Posts: 263 Thanks: 90  I did respond. Look again. Quote:
So why reply then? Hey, man, it's 5 am where I live and I'm replying to a math thread. Why do you think? Obviously: I can't sleep and I'm bored. Last edited by skipjack; February 27th, 2018 at 12:47 AM.  

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