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!