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.
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.