February 24th, 2018, 08:50 AM  #1 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,390 Thanks: 100  Limit of a Number
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? Last edited by zylo; February 24th, 2018 at 08:56 AM. Reason: Why? added 
February 24th, 2018, 09:48 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,009 Thanks: 1042 
This doesn't really make any sense as written. what is $S_{10}$ ? Are you supposing the existence of infinite digits? 
February 24th, 2018, 01:36 PM  #3 
Senior Member Joined: Sep 2016 From: USA Posts: 395 Thanks: 211 Math Focus: Dynamical systems, analytic function theory, numerics 
If I understand this correctly, then $S_{10}$ would be the integer 12345678910. If so, the limit doesn't exist, so claiming $S$ is equal to it is meaningless, as is asking whether or not the limit is a member of any set.
Last edited by skipjack; February 27th, 2018 at 12:19 AM. 
February 24th, 2018, 02:08 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,329 Thanks: 2454 Math Focus: Mainly analysis and algebra 
We seem to have all sorts of problems here. Not least that a number doesn't have a limit. Functions and sequences have limits (indeed sequences are functions with the natural numbers as their range). Limits aren't defined for any other objects that I can think of right now. SDK has hit the nail on the head for the implied sequence in your limit. Last edited by skipjack; February 27th, 2018 at 12:15 AM. 
February 25th, 2018, 12:14 PM  #5  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,240 Thanks: 885  Quote:
Last edited by skipjack; February 27th, 2018 at 12:23 AM.  
February 25th, 2018, 12:57 PM  #6 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,329 Thanks: 2454 Math Focus: Mainly analysis and algebra 
Er... Yeah. Ooops.

February 25th, 2018, 01:38 PM  #7  
Senior Member Joined: Aug 2012 Posts: 1,956 Thanks: 547  Quote:
In fact, $(S_n)$ is a subsequence of $(n) = 1, 2, 3, 4, 5, \dots$, which has the same limit. Last edited by skipjack; February 27th, 2018 at 12:21 AM.  
February 26th, 2018, 06:27 AM  #8  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,390 Thanks: 100  Quote:
And that $S \notin \{S_n\}_{n \in \mathbb N}$ Why? =========================== Sn, by definition, is a construction (artificial). If Sn is 12345.....n, what else could Sn be but 12345,....,10. If it's clearer, put a space between the numbers It's no different conceptually (see Maschke above) than Sn=n, for all n, and S=$\displaystyle \lim_{n \rightarrow \infty}$. Is S in {Sn}? In either case the answer is yes. Proof by contradiction: Assume S is not in {Sn}. Then there is an Sn for which n is a maximum. Contradiction. There is no maximum for n. It's really induction in disquise. If you don't accept induction as a definition of infinity, then "infinity" is not a mathematical term. If you don't precisely define your terms, it becomes a political discussion. For all (every) n iff induction defines infinity.  
February 26th, 2018, 06:51 AM  #9  
Senior Member Joined: Oct 2009 Posts: 428 Thanks: 144  Quote:
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 $n$.  
February 26th, 2018, 07:30 AM  #10  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,390 Thanks: 100  Quote:
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. Last edited by zylo; February 26th, 2018 at 07:39 AM.  

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