About uniformly continuity for bounded and continuous function Let $f:\left\{x\in\mathbb{R}^n\vert\parallel x\parallel<1\right\}\rightarrow\mathbb{R}$ be an one-to-one bounded continuous function.
I want to construct such $f$ which is not uniformly continuous.
In this case, I thought I can construct $f$ with a restriction $n=2$.
But I'm confused because $f$ is bounded so I can't use functions like $\frac{1}{x}$ on $(0,1)$.
To top that off, $f$ is even one-to-one so I gave up and now I'm writing this to get some help from you who is smarter than me.
Please give me some help to solve this problem.
Thanks.

For $n=1$, as pointed out by Vishal, there is no such function $f$.
For $n\geq 2$, such an $f$ doesn't exist either, but for a "trivial" reason: there is no continuous and one-to-one function at all from the open unit ball $B\subset\mathbb R^n$ into $\mathbb R$. One can see this as follows. Assume that $f:B\to\mathbb R$ is continuous and $1$-$1$. Then $I=f(B)$ is a nontrivial interval of $\mathbb R$ because $B$ is connected and $f$ is continuous and non-constant. Take any point $a\in B$ such that $f(a)$ is an interior point of $I$. Then $f(B\setminus\{ a\})$ has to be connected because $B\setminus\{ a\}$ is connected ($n\geq 2$) and $f$ is continuous. But $f(B\setminus\{ a\})$ is equal to $I\setminus \{ f(a)\}$ because $f$ is $1$-$1$, and this is not a connected set. So we have a contradiction.
