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February 26th, 2018, 08:49 AM   #11
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Quote:
 Originally Posted by zylo Or, if you prefer, Sn=n. What does the extended number system, a definition, or hyperreals have to do with this? EDIT As I wrote in a previous post, my answer is yes. Assume S is not in {Sn}. Then there is an Sn for which n is a maximum. Contradiction. There is no maximum for n.
OK then, so take $S_n=n$ as you suggested. Take
$$S = \lim_n S_n$$
(whatever this means)
For which exact natural number do we have $S=S_n$ then exactly?

February 26th, 2018, 09:21 AM   #12
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Quote:
 Originally Posted by Micrm@ss OK then, so take $S_n=n$ as you suggested. Take $$S = \lim_n S_n$$ (whatever this means) For which exact natural number do we have $S=S_n$ then exactly?
S is not an exact natural number, but {Sn} includes S because n is not bounded.

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,

Last edited by zylo; February 26th, 2018 at 09:29 AM.

February 26th, 2018, 10:04 AM   #13
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Quote:
 Originally Posted by zylo S is not an exact natural number, but {Sn} includes S because n is not bounded. 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,
OK, so this is weird. A natural number to me is something of the form 2, 13, 100000, and they all have such an explicit notation. You can prove all natural numbers have an explicit notation, and the only reason you can't give it is because it's too long.
So what would a natural number be if it has no exact explcit notation?

February 26th, 2018, 10:23 AM   #14
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Quote:
 Originally Posted by Micrm@ss OK, so this is weird. A natural number to me is something of the form 2, 13, 100000, and they all have such an explicit notation. You can prove all natural numbers have an explicit notation, and the only reason you can't give it is because it's too long. So what would a natural number be if it has no exact explcit notation?
There is no last natural number, just as there is no last Sn.

I can define Sn for all n, so Sn has a definition for all n.

So $\displaystyle \lim_{n \rightarrow \infty}$ defines S.

If I could only define Sn for a finite number of terms, I couldn't define S.

EDIT:
Ex of infinite sequence of binary digit: 101010......10 to n places and let n $\displaystyle \rightarrow \infty$

Bottom line is, the only mathematical (hard) definition for $\displaystyle$\displaystyle \infty$$I can think of is induction, ie, for all n. Just out of curiosity, what do you think "an infinite sequence of binary digits" means, and how would you give an example of one? Last edited by zylo; February 26th, 2018 at 10:39 AM.  February 26th, 2018, 10:26 AM #15 Senior Member Joined: Oct 2009 Posts: 733 Thanks: 246 Wait, are you talking about something like$$S=123456789101112....$$and then an infinite number of digits? Your post seems to suggest this. You do realize that those things won't be natural numbers though, right? February 26th, 2018, 11:18 AM #16 Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 Quote:  Originally Posted by Micrm@ss Wait, are you talking about something like$$S=123456789101112.... and then an infinite number of digits? Your post seems to suggest this. You do realize that those things won't be natural numbers though, right?
Yes. I never said it was a natural number.

Quote:
 Originally Posted by zylo S = $\displaystyle \lim_{n\rightarrow \infty}$ 123....n $\displaystyle S_{n}$ = 123....n for all n. Is S a member of {$\displaystyle S_{n}$}? Why?
S is a specific, unique, unending sequence. As such, it is a member of {Sn}

Last edited by zylo; February 26th, 2018 at 11:22 AM.

February 26th, 2018, 11:48 AM   #17
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Quote:
 Originally Posted by zylo S is a specific, unique, unending sequence. As such, it is a member of {Sn}
None of the members of $\{S_n\}$ are "unending". They are all by definition finite, so $S \not \in \{S_n\}$.

February 26th, 2018, 12:17 PM   #18
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Quote:
 Originally Posted by zylo Yes. I never said it was a natural number.
OK, but each $S_n$ is a natural number. So if $S$ is in $\{S_n\}$, wouldn't that make $S$ a natural number?

(this is essentially the same as v8archie's post)

 February 26th, 2018, 01:12 PM #19 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 I said Sn was not a natural number. I simply meant it to be an example of something constructed, or defined by, natural numbers. If it makes it easier, put spaces between. S$\displaystyle _{11}$ = 1 2 3 4 5 6 7 8 9 10 11 $\displaystyle \lim_{n \rightarrow \infty}$ Sn =1 2 3,......,12578, .......... If none of {Sn} are unending, then there has to be a maximum value of n for Sn. There isn't. Last edited by skipjack; February 27th, 2018 at 01:37 AM.
February 26th, 2018, 02:21 PM   #20
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Quote:
 Originally Posted by Micrm@ss True, but I wonder if the extended real numbers are the best setting to see this. I think it is worth checking out the hyperreals. The hyperreals are by definitions all the sequences in $\mathbb{R}$ modulo some equivalence relation. In this system, $S_n$ does not have the same limit as $n$.
Very interesting comment that I'd like to understand with your help. I dove into the hyperreals a couple of years ago. I read Terence Tao's articles on NSA including his awesome explanation of ultrafilters as voting systems. I know what a nonprincipal ultrafilter is. I read Lightstone's paper on his notation for hyperreals and I read a few other papers besides. There was a moment when I thought I had it all straight in my mind but apparently that clarity is gone now.

I mention this so that you can calibrate the state of my knowledge. Can you please explain how $(S_n)$ has a different limit than $(n)$ in the hyperreals or hyperintegers? Thanks much.

Last edited by Maschke; February 26th, 2018 at 02:33 PM.

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