HW 6 Challenge 2 a,b

Consider the function f(x)=\frac{1}{x}.

We will consider the domains (0,1) and [1,\infty).

A) To prove that f(x)=\frac{1}{x} is not uniformly continuous on (0,1), we need to show that for some \epsilon_\circ>0, there exist sequences (x_n), (y_n)\in (0,1) such that |x_n-y_n|\rightarrow 0, but |f(x_n)-f(y_n)|\geq \epsilon_\circ.

Let (x_n):=\{x_n\in (0,1): x_n=\frac{2n+1}{2^n} and (y_n):=\{y_n\in (0,1): y_n=\frac{2n}{2^n} for n\in \mathbb{N}. Notice that |x_n-y_n|=|\frac{2n+1}{2^n}-\frac{2n}{2^n}|=|\frac{1}{2^n}|\rightarrow 0. Yet |f(x_n)-f(y_n)|=|\frac{2^n}{2n+1}-\frac{2^n}{2n}|=|\frac{2^n}{2n(2n+1)}|>0 for all n\in \mathbb{N}. Hence by the archimedean property we can choose 0<\epsilon_\circ<|\frac{2^n}{2n(2n+1)}|. Thus f(x)=\frac{1}{x} is not uniformly continuous on (0,1).

B) To prove that f(x)=\frac{1}{x} is uniformly continuous on [1,\infty), we need to show that for every \epsilon>0, there exists a \delta>0 such that for all x,c\in [1,\infty), |x-c|<\delta implies |f(x)-f(c)|<\epsilon.

Let \epsilon>0 be arbitrary. Choose a \delta<\epsilon. Let c\in [1,\infty) be arbitrary and consider x\in [1,\infty) with |x-c|<\delta. Thus |f(x)-f(c)|=|\frac{1}{x}-\frac{1}{c}|=|\frac{x-c}{xc}|\leq |x-c|<\epsilon since we have xc\geq 1.

Hence f(x)=\frac{1}{x} has uniform continuity on [1,\infty) because c was arbitrary.

Posted on by Elaine Wissuchek
Categories: Challenges

3 Responses

  1. Jeremy LeCrone says:
    October 24, 2017 at 12:05 pm

    https://richmond.qualtrics.com/jfe/form/SV_0f8VdboeDIysNX7

  2. Jonathan Rodriguez says:
    October 28, 2017 at 2:09 pm

    I think the way you tried to apply the Archimedean Property to this problem was clever. Unfortunately in class we found out that this method would not provide that single epsilon we were looking for. However, the correction you provided in class was very solid.

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