# Interactive Real Analysis - part of MathCS.org

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## Example 6.2.8(a):

The function f(x) = 1 / x is continuous on (0, 1). Is it uniformly continuous there ?
It helps to look at the graph of the function:
As the - interval 'slides' up the positive y-axis, the corresponding -interval on the x-axis gets smaller and smaller. That indicates that the function is not uniformly continuous - but it is of course not a proof. So:

Take any = 1. Does there exist any such that

if | t - s | < then | f(t) - f(s) | < = 1 ?

The basic idea is easy: since | f(t) - f(s) | = | 1/t - 1/s | = | s - t | * 1 / | st |, we can see that this might approach infinity if s and t approach zero, and therefore will be bigger than any chosen . All we have to do now is formalize this proof.

Assume there exists such a . Without loss of generality we may assume that < 1 (why ?). Then let s = t + / 2 and set t = / 2. We have (assuming that s, t are positive):

| f(t) - f(s) | = | 1/s - 1/t | = | (t - s) / st | = > 1

But now, no matter what < 1 is we can make | f(t) - f(s) | > 1. Therefore, the function is not uniformly continuous.

This proof, loosely speaking, depends on the fact that after simplification | f(t) - f(s) | goes to infinity if s and t approach zero. That is exactly the situation as described in the above picture.

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