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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...."? 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. 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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