# Prove $1/x$ is not uniformly continuous

$$f: (0,+\infty) \to (0,+\infty)$$ $$f(x) = 1/x$$, prove that f is not uniformly continuous.

Firstly, I negated the definition of uniform convergence obtaining:

$$\exists \epsilon > 0$$ s.t. $$\forall \delta > 0$$ with $$|x-y| < \delta$$ & $$x,y \in (0, + \infty)$$ and $$|f(x) - f(y)| = \left|\dfrac{x-y}{xy}\right| \geq \epsilon$$

so I choose $$\epsilon = 1$$ and $$x = \delta/2 \in (0,+\infty)$$ and $$y = \delta /4 \in(0,+\infty)$$ so $$|x-y| = \delta/2 < \delta$$ and $$|f(x) - f(y)| = |2/\delta|$$ How do I show that this is greater than or equal to epsilon?

The condition you want is $|2/\delta|\ge \epsilon = 1$, which holds true as long as $|\delta|\le 2$. So you're set for small $\delta$. Can you see a way to manage the case when $\delta>2$ accordingly?

• I'll give it a go - but is there something specifically wrong with mine?
– Warz
Jan 20, 2014 at 19:40
• @Warz See my edit. Jan 20, 2014 at 19:48
• Could we not just set $\epsilon = 1/\delta$ then we would still get $|2/\delta| > 1/\delta = \epsilon$?
– Warz
Jan 20, 2014 at 19:53
• No, you don't get to set $\epsilon$ in terms of $\delta$. Jan 20, 2014 at 19:55
• hmm, I don't really see what we could do then - I always thought of delta as small so didn't really consider the other case.
– Warz
Jan 20, 2014 at 19:56

Does the function take Cauchy sequences to Cauchy sequences? What does it take the sequence of numbers $x_n = n^{-1}$ to ?