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Is $f(x)=\frac{1}{x}$ uniformly continuous then x $\in (0,1)$?

I think that it's not uniformly continuous so I am trying to prove that there exists an epsilon>0 for all deltas>0 and there exist x,y such that

$|f(x)−f(y)|≥ϵ$ if $ |x−y|<δ$

I started by choosin epsilon=1 . Then:


Now, I think, I need to choose such values of x and y expressed through delta such that the equality above would be greater or equal than one, but I am having trouble thinking of such values. Is my approach any good?

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up vote 2 down vote accepted

We can see that it's not uniformly continuous, because a function that is uniformly continuous on an interval must be bounded, whereas the function is not. This alone is actually enough for a proof. If we want to do it from first principles however, we can do the following.

let $x = \delta$, $y = \frac{\delta}{2}$. Then $|x-y|< \delta$, and

$$|\frac{1}{x} - \frac{1}{y}| = \frac{|x-y|}{xy} = \frac{\frac{\delta}{2}}{\frac{\delta^2}{2}} = \frac{1}{\delta} > 1$$

Since we must have $\delta < 1$.

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Why delta must be less than 1? Shouldn't this proof work for every delta? – user1242967 Dec 11 '12 at 19:10
@user1242967 It does, since we are working in $(0,1)$, $x, y\in (0,1) \implies |x-y| < 1$. Even if we weren't, since we are trying to prove $f$ is not uniformly continuous, we don't have to prove "...for every $x,y$ such that $|x-y|< \delta$..." but only "...there exists $x,y$ such that $|x-y| < \delta$..." So if it is true for some small $\delta$, it must also be true for any $\alpha > \delta$ since $|x-y| < \delta \implies |x-y| < \alpha$. – Tom Oldfield Dec 11 '12 at 19:15

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