# Math Help - D bounded => f is bounded on D

1. ## D bounded => f is bounded on D

I need some help with a proof
Assume $f\to \mathbb{R}" alt="f\to \mathbb{R}" /> is uniformly continous on D.
Show that D being bounded implies f is bounded on D.

All I can think of is to assume f is unbounded and show that it leads to a contradiction, but not sure how to proceed with the proof.
any suggestions? tHanks

2. Precompactness is the key. Let $f$ be uniformly continuous on $D\subseteq \mathbb{R}^n$. You get d>0 such that the oscilation of $f$in each ball of radius d is smaller than 1. Every bounded subset in $\mathbb{R}^n$ is precompact, hence you can find $x_1,\ldots, x_n\in D$ such that $D\subset\cup_{i=!}^{n}B(x_i,d)$. Now you can easily get a bound for $|f|$.