Prove that the function defined by $f(x) = \sin(1/x)$ is not uniformly continuous on the interval $(0,1)$.
Hint: Consider for example $x = 1/2nπ$ and $y = 1/[(2n+1/2)π]$
I have an answer with let $ε = 1/2$
we must have $|f(x)-f(y)|< ε =1/2$
$f(x)= \sin(1/x)=\sin(2nπ)$
and $f(y)= \sin(1/y)= \sin(2nπ+π/2)$
$|f(x)-f(y)|=|\sin(2nπ)-\sin(2nπ+π/2)|=|0-1|=1 > 1/2=ε$
so $f(x)$ by contradiction is not uniformly continuous on $(0,1)$
is this going in the right direction. Im really unsure if i have missed an important part! thanks for your help!
Adding this:
so i have $|x-y|<δ$ let $δ>0, |x-y|=|1/2nπ – 1/(2n+1/2)π|=|[π/2]/[2nπ(2nπ+π/2)| < δ$ so we have that when $n$ tends to infinity $|x-y|$ tends to $0$
is this complete then? Sorry for the confusing symbols!
Best Answer
Your on the right track. You just need to write it in such a way that the reader can be sure that there's only one thing you could mean. The point is that there is no $\delta>0$ such that for all $x,y$ at distance $<\delta$ from each other you have $|f(x)-f(y)|<1/2$. That's the same as saying that for every $\delta>0$ there exist $x,y$ such that $|x-y|<\delta$ and $|f(x)-f(y)|\ge1/2$. Your way of finding $x$ and $y$ is correct.