Thread: Uniform Continuity

|x|

Originally Posted by lovesmath

Prove that the function is uniformly continuous on the set of real numbers.
f(x)=|x|
I think I have to use the reverse triangle inequality, but I am not sure how to do it.

There you have it.

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.

Originally Posted by lovesmath

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.

What have you tried so far?

-Dan

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.

Originally Posted by lovesmath

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.

The key to this is the fact that is bounded.
That means that such that .

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.