Let $X, Y, Z$ be metric spaces. $f: X \rightarrow Y$ is called uniformly continuous if
$\forall \epsilon 0 \quad \exists \delta 0 \quad \forall x, y \in X : d_{x}(x,y) \leq \delta \Rightarrow d_{Y}(f(x),f(y))\leq \epsilon $
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b) $f(x):=x^{2}$ is continuous, but not uniformly continuous on $\mathbb{R}$

Question

Let $X, Y, Z$ be metric spaces. $f: X \rightarrow Y$ is called uniformly continuous if
     $\forall \epsilon > 0 \quad \exists \delta > 0 \quad \forall x, y \in X : d_{x}(x,y) \leq \delta \Rightarrow d_{Y}(f(x),f(y))\leq \epsilon $
   Show:

   b) $f(x):=x^{2}$ is continuous, but not uniformly continuous on $\mathbb{R}$

anonymous · Answer

you can proove  the not uniformly continuous part by finding counterexample. Try to take: $$\epsilon=1$$ and work out the rest from definition.

anonymous · Answer

myko thank you very much, but i need complete solution because about 40 min ago i should it deliver to my mentor... i mean i dont have time try to understand it and solve....

anonymous · Answer

any help would be apreciated

anonymous · Answer

then just look maybe here: http://answers.yahoo.com/question/index?qid=20101103061142AAP8pTE

anonymous · Answer

ok thanks

anonymous · Answer

there it says$$[1,\infty]$$ but we have in question about $$\mathbb{R}$$ is it much difference?

anonymous · Answer

If it isn't uniformly continuous on a subset of $\mathbb R$ then it certainly isn't uniformly continuous on $\mathbb R$.

anonymous · Answer

ok thx

anonymous · Answer

Take
$$
x_n = n + \frac 1 n\
y_n= n - \frac 1 n\
x_n - y_n= \frac 2n\
x_n^2 - y_n^2= 4\
\lim_{n\to \infty} x_n -y_n=0 \ \text { but }
\lim_{n\to \infty} x_n^2 -y_n^2=4
$$
So f(x) is not uniformly continuous.