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I have a quick question on the excerpt of Theorem 2.40 of Baby Rudin. How would I get "If n is so large that $2^{-n}\delta<r$...."? I think that to get such an 'n' has something to do with exercise on Chapter 1 which is related to logarithm which I have not completed yet. I was trying to rephrase the excerpt as the following lemma: There exists $n$ such that $2^{-n}\delta<r$. Proof:Let $E=\{2^n:n\in \mathbb{N}\}$. Suppose not. Then $2^n \le \frac{\delta}{r}$. Since $E\subset \mathbb{R}$, $E$ is bounded above, therefore $\sup E= \alpha$ exists, i.e. $2^n \le \alpha$ for all $n$. But I could not really proceed to get a contradiction though.

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You want to have $2^n>\dfrac{\delta}r$; taking logs, this is equivalent to $n\ln 2>\ln\delta-\ln r$, or $n>\dfrac{\ln\delta-\ln r}{\ln 2}$. The fact that you can always choose such an $n$ comes from the Archimedean property of $\Bbb R$.

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Thanks Brian. I am just not sure if I can actually use natural logarithm, unless this is found somewhere in Exercise on Chapter 1 which has something to do with logarithm. –  Daniel Jun 11 '12 at 7:11
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To finish the lemma, note that $\alpha=\sup E$ is the least upper bound for $E$, therefore $\alpha/2<\alpha$ is not an upper bound for $E$. Thus there is some $n$ such that $2^n>\alpha/2$, so $2^{n+1}>\alpha$. This contradicts the fact that $\alpha$ is the least upper bound for $E$.

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Yes thanks a lot Alex! I think your answer comes closest to what I want since I have not actually completed the exercise about logarithm of Chapter 1. –  Daniel Jun 11 '12 at 7:09
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