I have to prove that the function $f(x)=\frac{1}{x}$ on $(0,\infty)$ is not uniformly continuous (for the definition of uniform continuity see here).
I negated the definition of uniform continuity getting:
$$ \exists \epsilon>0 \ \text{s.t.} \ \forall \delta>0 \ \text{I can always find} \ x,y \in(0,\infty) \ \text{s.t.} \mid x-y\mid<\delta \ \text{but} \ \mid f(x)-f(y)\mid \geq \epsilon. $$
So I chose $\epsilon=1$. Set $y=\frac{x}{2}$. Then I have to find an $x$ such that this holds: $\mid x-y\mid = \frac{x}{2} <\delta$ & $\mid f(x)-f(y)\mid = \frac{1}{x} \geq 1$, which is equivalent to $ x<\min [2\delta,1] $.
Is this argument valid? My concern is if I'm allowed to choose $x$ depending on $\delta$. Thanks!
Best Answer
It is correct. You can choose x depending on $\delta$ because $x$ is quantified before $\delta$.
Alternatively, we have $$1/n - 1/(n+1) \to 0$$ but
$$f(1/n) -f(1/(n+1)) = n - (n+1) \to -1 $$
And by the equivalent sequence characterisation of uniform continuity, we conclude that $f$ cannot be uniformly continuous.