# Uniform Continuity: Is this a good proof?

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
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• 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.