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Show that $f(x)=1/x^2$ is not uniformly continuous one the set $(0,1]$

Using the Sequential Criterion for Nonuniform Continuity - which states that

a function $f:A \rightarrow $ R fails to be uniformly continuous on A iff there exists a particular $\epsilon_0$>0 and two sequences ($x_n$) and ($y_n$) in A, satisfying $|x_n -y_n| \rightarrow 0$, but $|f(x_n) - f(y_n)|\ge \epsilon_0$

I would say:

Take ($x_n$) = $\frac{1}{n}$ and ($y_n$)=$\frac{1}{n^2}$. Obviously $|\frac{1}{n} - \frac{1}{n^2}| \rightarrow 0$, but $|f(x_n) - f(y_n)| = |n^2 - n^4| \ge 12 $, for example for n $\ge$ 2

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Uniformly continuous functions take Cauchy sequences to Cauchy sequences. That seems as the easiest way to show that your function isn't uniformly continuous. –  kahen Oct 20 '12 at 17:34
Or: If $f$ is uniformly continuous on a bounded domain, then $f$ is bounded. Your $f$ is not bounded, but $(0,1]$ is. –  Hagen von Eitzen Oct 20 '12 at 18:11
I've noticed that you have asked 5 questions in the last 24 hours. I wanted to make sure that you are aware of the quotas 50 questions/30 days and 6 questions/24 hours, so that you can plan posting your questions accordingly. (If you try to post more questions, Stackexchange software will not allow you to do so.) For more details see meta. –  Martin Sleziak Oct 21 '12 at 13:41
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1 Answer

up vote 1 down vote accepted

This CW answer is intended to remove this question from the Unanswered queue.

There is nothing wrong with your argument involving the Sequential Criterion.

As pointed out in the comments, you could also have approached this by:

  • exhibiting a Cauchy sequence $(x_n)_n$ such that $(f(x_n))_n$ is not Cauchy;
  • showing that the range of $f$ is not bounded, while its domain is.
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