Uniformly continuous in uniform space

Question

Let $(\Bbb R,U)$ be the uniform space induced by the usual metric space $(\Bbb R,d)$ .Show in details that the function $f:(\Bbb R,U)\to (\Bbb R,U)$ , $f(x)=x^3$ is homemorphism, but not uniformly continuous?

Answer 1

Suppose that $f(x)=x^3$ is uniformly continuous w.r.t. the standard uniformity $\mathcal{U}_d$ induced by $d(x,y)=|x-y|$ on $\Bbb R$, i.e. $U \in \mathcal{U}_d$ iff $\exists r>0: U(d,r):=\{(x,y): d(x,y) < r\} \subseteq U$.

And recall that $f: (\Bbb R, \mathcal{U}_d) \to (\Bbb R, \mathcal{U}_d)$ is uniformly continuous iff $$\forall U \in \mathcal{U}_d: (f \times f)^{-1}[U] \in \mathcal{U}_d$$

In particular there must be some $\delta>0$ such that

$$U(d,\delta) \subseteq (f \times f)^{-1}[U(d,1)]$$

which implies that for that $r>0$ we have

$$\forall x,y \in \Bbb R: d(x,y) < \delta \to d(f(x), f(y)) < 1$$

and this is easily refuted (so that no such $\delta$ can exist):

(adapting from this post:)

Choose $x$ large enough so that $\frac{3\delta x^2}{2}>1$; for example $x=\sqrt{\frac{2}{3\delta}}+1$ works. Now take $y=x+\frac{\delta}{2}$; this satisfies $|x-y|=\frac{\delta}{2}<\delta$. Hence we should have $|f(x)-f(y)|<1$. Instead we have $|f(x)-f(y)|=|f(x+\frac{\delta}{2})-f(x)|=|(x+\frac{\delta}{2})^3-x^3|=\left|3x^2\frac{\delta}{2}+3x\frac{\delta^2}{4}+\frac{\delta^3}{8}\right|\ge \left|3x^2\frac{\delta}{2}\right|>1$.

So for such $(x,y)$ we have $(x,y) \in U(d,\delta)$ while $(f \times f)(x,y) \notin U(d,1)$. So $f$ is not uniformly continuous.

That $f$ is continuous and has a continuous inverse $g(x)=\sqrt[3]{x}$ is easy to check, so $f$ is a plain topological homeomorphism.

Answer 2

Clearly the function $f(x)=x^3$ is NOT UNIFORMLY CONTINUOUS but continuous as every polynomial is continuous. Now to show bijectivity note that : $f$ is one one as $a\ne b\implies a^3\ne b^3,\forall a,b\in\Bbb R$, and surjective since for every $x\in \Bbb R$ set $y=x^\frac{1}{3}$ so that $f(y)=y^3=x$ i.e. $f$ is onto and hence bijection. Now the inverse map $f^{-1}(y)=y^\frac{1}{3}$ is also continuous (left as an exercise) which means $f$ is an open map.

Thus $f$ is an open bijective continuous map i.e. a homeomorphism.

Hope this works.