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 May 18th, 2018, 10:29 AM #1 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 Limits Observation Let {$\displaystyle a_{n}$} be a sequence. $\displaystyle \lim_{n\rightarrow \infty}a_{n}=L$ if, given $\displaystyle \epsilon$, N exists st $\displaystyle |a_{n}-L| < \epsilon$ for all n > N but there are an infinite number of sequences which satisfy this condition, NO MATTER WHAT $\displaystyle \epsilon$ is, namely all sequences which are the same up to $\displaystyle a_{N}$, so you can never arrive at a unique sequence for L. This has implications for decimal representation of numbers. I haven't thought through implications for Cauchy sequences and completeness. Offhand, it looks like the only argument for completeness is Dedekind cuts. May 18th, 2018, 11:42 AM   #2
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 Originally Posted by zylo Let {$\displaystyle a_{n}$} be a sequence. $\displaystyle \lim_{n\rightarrow \infty}a_{n}=L$ if, given $\displaystyle \epsilon$, N exists st $\displaystyle |a_{n}-L| < \epsilon$ for all n > N but there are an infinite number of sequences which satisfy this condition, NO MATTER WHAT $\displaystyle \epsilon$ is, namely all sequences which are the same up to $\displaystyle a_{N}$, so you can never arrive at a unique sequence for L.
No, the inequality you give is only required for n> N. What happens for n "up to N" is irrelevant. In any case, yes, given a number L there exist infinitely many sequences that converge to L.

Quote:
 This has implications for decimal representation of numbers. I haven't thought through implications for Cauchy sequences and completeness. Offhand, it looks like the only argument for completeness is Dedekind cuts.
What implications are you talking about? I presume that you mention "Cauchy sequences" because one way to define the real numbers in terms of rational numbers is to define a real number as 'an equivalence class of Cauchy sequences of rational numbers', the equivalence relation being {a_n} is equivalent to {b_n} if and only if the sequence {a_n- b_n} converges to 0. The whole point of taking an equivalence class is because there are an infinite number of sequences that converge to a given L. May 18th, 2018, 11:54 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,656 Thanks: 2634 Math Focus: Mainly analysis and algebra There has never been any suggestion that there was a unique sequence for each $L$. That's trivial. \left.\begin{aligned}\lim_{n \to \infty} a_n &= L \\ \lim_{n \to \infty} b_n &= 0 \end{aligned}\right\} \implies \lim_{n \to \infty} (a_n +b_n) = \lim_{n \to \infty} (a_n -b_n) = L Tags limits, observation 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 mrtwhs Calculus 1 March 1st, 2016 02:31 PM jasonjason Computer Science 0 July 26th, 2013 01:47 AM soroban New Users 3 September 10th, 2012 12:50 PM MyNameIsVu Applied Math 9 July 17th, 2009 10:08 PM kaushiks.nitt Number Theory 31 June 19th, 2009 07:26 PM

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