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So 1=0.999... can be proved by arguing that the only real number can be put between 1 and 0.999... is zero due to the fact that the only infinitesimal in real number is zero.

We can also write the statement as: $$ (\exists c \in \Bbb R)(1-0.999\cdots=c) \Rightarrow (c=0) $$

Also, I saw another proof in wiki by expressing 0.999.. in an infinite geometric series and calculate its limit to be 1. i.e. $$ 0.999\ldots = \lim_{n\to\infty}0.\underbrace{ 99\ldots9 }_{n} = \lim_{n\to\infty}\sum_{k = 1}^n\frac{9}{10^k} = \lim_{n\to\infty}\left(1-\frac{1}{10^n}\right) = 1-\lim_{n\to\infty}\frac{1}{10^n} = 1 \, - \, 0 = 1. $$

However, the epsilon-delta definition of limit can only ensure that the difference between the summation and its limit to be smaller than smaller than a positive real number $ \epsilon $ . Under such definition, the infinite geometric series above has a limit of 1 but it is actually never reaching 1. The best we can ensure that the infinite geometric series is $ \epsilon $ away from 1.

So my question is:

Under rigorous definition of limit by the epsilon-delta definition and real number system, is it valid to prove 0.999...=1 by infinite geometric series?

Supplement:

So some of you mentioned that the limit of such infinite geometric series is 1. According Epsilon-delta definition of limit, limit is

$$ \lim_{x \to c} f(x) = L \iff (\forall \varepsilon > 0)(\exists \ \delta > 0) (\forall x \in D)(0 < |x - c | < \delta \ \Rightarrow \ |f(x) - L| < \varepsilon) $$

So we can only find an $ x $ such that $ |f(x) - L| < \varepsilon $. They never equal.The limit of f(x) equals to L doesn't mean f(x) itself equals to L when there exist an $ x $ such that $ 0 < |x - c | < \delta $ .

Ken T
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1 Answers1

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Not only is it valid, but given that the limit of the series is actually the very definition of what $0.999\ldots$ means, any rigorous proof of $0.999\ldots=1$ is ultimately either itself a prove of that limit, or in some way depends on the proof of that limit.

celtschk
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  • This has not answered my question. This simply asks me to believe. – Ken T Mar 23 '17 at 11:50
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    @KenT: It answers the question because the definition is the only thing that gives the symbols a meaning. Without using the definition, something like $0.9999\ldots$ is not even a number. It's just a meaningless combination of symbols, until you apply the definition of what the symbols mean. And the definition is exactly the series, plain and simple. – celtschk Mar 28 '17 at 09:25