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- Nov 17th 2012, 02:44 PMlovesmathUniform Continuity
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- Nov 17th 2012, 02:50 PMPlatoRe: Uniform Continuity
- Nov 17th 2012, 04:16 PMlovesmathRe: Uniform Continuity
Here is an additional question.

Suppose f is a uniformly continuous function on a bounded set D contained in the set of real numbers. Prove that f(D) is a bounded set. - Nov 17th 2012, 05:15 PMtopsquarkRe: Uniform Continuity
- Nov 18th 2012, 05:52 PMlovesmathRe: Uniform Continuity
I know I can set Epsilon=1. There exists a delta>0 such that x,y are elements in D and |x-y|<delta implies that |f(x)-f(y)|<E. Then |f(x)-f(y)|<1.

Not sure what to do next. - Nov 19th 2012, 04:12 AMPlatoRe: Uniform Continuity
The key to this is the fact that $\displaystyle D$ is bounded.

That means that $\displaystyle \exists M>0$ such that $\displaystyle D\subset [-M,M]$.

In your text material you should have a proof that a continuous function is bounded on a closed interval of real numbers. This new proof follows that proof very closely.