$\displaystyle f(x)=1/x$ how would you show this is continuous on $\displaystyle A=[1,\infty)$ but not on $\displaystyle B=(0,\infty)$?
I assume you mean uniformly continuous
First note that 1/x - 1/y = y-x/xy
For the first part:
For x, y in B, we have x,y>1 so 1/xy < 1
Thus, for all x,y in B lf(x) - f(y)l = l1/x - 1/yl = lx-yl/lxyl < lx-yl
and uniform continuity is not hard to show (let delta = epsilon).
For the second part, you have to show that there exists an epsilon > 0, you can find an x, y small enough such that their difference is greater than epsilon.
To prove that 1/x is not uniformly continuous on (0,$\displaystyle \infty$) we must show that:
There exists ε>0 such that for all δ>0 there exists x,wε(0,$\displaystyle \infty$) such that:
|x-w|<δ and |f(x)-f(w)|$\displaystyle \geq\epsilon$.
Let ε=2.
Take 0<δ<1 .
Choose x= δ and w= δ/4 then we have:
|x-w| = |δ-$\displaystyle \frac{\delta}{4}|= \frac{3\delta}{4}<\delta$
AND
|f(x)-f(w)| = $\displaystyle |\frac{1}{\delta}-\frac{4}{\delta}|=\frac{3}{\delta}\geq 3\geq\epsilon$.
Hence 1/x is not uniformly continuous on (0,$\displaystyle \infty$)
as aside note, $\displaystyle f(x)=\frac1x$ is uniformly continuous for each $\displaystyle [\alpha,\infty)$ where $\displaystyle \alpha>0.$
the proof follows easily since given $\displaystyle \epsilon>0$ and by taking $\displaystyle \delta=\alpha^2\epsilon$ we have $\displaystyle \left| \frac{1}{x}-\frac{1}{x_{0}} \right|=\frac{\left| x-x_{0} \right|}{x\cdot x_{0}}\le \frac{1}{\alpha ^{2}}\left| x-x_{0} \right|<\frac{1}{\alpha ^{2}}\cdot \delta =\frac{1}{\alpha ^{2}}\cdot \alpha ^{2}\epsilon =\epsilon.\quad\blacksquare$