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 May 18th, 2018, 11:29 AM #1 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 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, 12:42 PM   #2
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Quote:
 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, 12:54 PM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,511 Thanks: 2514 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

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