Is there exist an $f:X\to \mathbb{R}$ where $\sup f(X)= +\infty$ and $f$ is uniformly continuous on $X$ where X is bounded? I think there should not be existing such $f$ as the change of the value of $f$ too quick
Is there exist such $f$
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0@AlexYoucis I think (s)he probably means $X$ a bounded subset of $\mathbb{R}$, not $f$ bounded – 2012-12-09
3 Answers
As others have said there is no such $f$ when $X\subset{\mathbb R}$. But in an arbitrary metric space $X$ you may have such $f$'s. Here is an example:
Let $X:={\mathbb R}$ and provide $X$ with the metric $d(x,y):={|x-y|\over 1+|x-y|}\ .$ Then all distances in $X$ are $<1$, so $X$ is "bounded". But for $|x-y|\ll 1$ the new distance is pretty much the usual one. It follows that the unbounded function $f(x):=\log\bigl(1+|x|\bigr)$ is locally Lipschitz continuous with Lipschitz constant $1$, whence uniformly continuous.
Assuming, as per Deven Ware's comment, you mean that $X$ is bounded. Then, what you say is true. Namely, since $f$ is uniformly continuous and $\mathbb{R}$ is complete, it's a common fact that you can extend $f$ to a map $\widetilde{f}:\overline{X}\to\mathbb{R}$ by merely defining $\widetilde{f}(x)=\lim f(x_n)$ where $x_n$ is a sequence in $X$ converging to $x$. But, then $\widetilde{f}$ is continuous on the bounded and closed set $\overline{X}$, and thus $\widetilde{f}$ and thus $f$ is bounded.
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0I'm sorry, I don't know what you mean. What I'm saying is that you are correct. Any uniformly continuous real function on a bounded subset of $\mathbb{R}$ is bounded (of course, more is true if you look closely at my proof). – 2012-12-09
Since $X$ is bounded there exists some diameter $d \geq 0$ such that $|x - y| \leq d$ for all $x, y \in X$. Let $x_0, x_1, \dotsc$ be a sequence in $X$ such that $f(x_{n+1}) \geq f(x_n) + 1$. Then $|f(x_a) - f(x_b)| \geq 1$ for all indices $a\neq b$. Take some integer $N > 0$. Then there are indices $a, b \leq N$, $a \neq b$ such that $|x_a - x_b| \leq \tfrac{d}{N}$. Since $N$ can be chosen arbitrarily large, $f$ is not uniformly continuous on $X$.