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This is a proof from Royden's Real Analysis and there is a part that I do not understand( Underlined in red). Any explanation is much appreciated.

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To put it in simply, consider a point $x_0$ in the set $(a,b) \sim C$. Now take a neighborhood about $x_0$ and notice that if you take a neighborhood small enough you can ensure that no points $q_1 ,q_2, \ldots ,q_{n}$ are in this neighborhood. Therefore if $x$ is a point in the said neighborhood then $|f(x_0)-f(x)|$ will be at most the tail end of the series $\sum \frac{1}{2^n}$ from the point $\frac{1}{2^{n+1}}$ onward. And this tail end sum can be shown to be less than $\frac{1}{2^n}$

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It is critical to note that $n$ is chosen before the interval $I$ is selected. There is a typo in that the $q_n$ you underlined should be $q_k$. Given $x_0$ and $n$, there are only finitely many $q_k$ with $k \le n$, so one of them is closest to $x_0$. You can't have an infinite sequence that gets closer and closer to $x_0$ because the sequence isn't infinite. You can then choose $I$ to be small enough it does not include any of the $q_k$.

This is leading to an $\epsilon-delta$ proof. If you give me an $\epsilon$, I need to find a $\delta$ such that ... Instead of finding $\delta$ algebraically, I find $n$ such that if there aren't any $q_k$ with $k \le n$, the jumps in the interval will be small enough.

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    $\begingroup$ Yes I knew it had to be a typo! This book is full of typos! $\endgroup$
    – user81883
    Dec 5 '16 at 4:33

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