Real Analysis True/False Statement?

Question

Statement: The function f(x) = x*x (x squared) is uniformly continuous on two of the following three sets: (0,1), Z, R. Z is the set of all integers and R is the set of all real numbers.

My answer: True

Answer 1

The first two of the above answers are wrong. You are correct, this is true. x² is uniformly continuous on (0, 1) since ∀ε>0, ∃δ=ε/2 s.t. |x-y|<δ ⇒ |x²-y²| = |x+y||x-y| < 2δ = ε. It is uniformly continuous on Z since ∀ε>0, ∃δ=1 s.t. |x-y|<δ ⇒ |x²-y²|<ε (note that this implication is vacuously true, since on Z, |x-y|<1 never happens). It is not uniformly continuous on Z, because although it is continuous, there is no δ which suffices for all x, y such that |x-y|<δ (intuitively, this is because the derivative of x² is not bounded on R).

Incidentally Gabe, stop deleting your questions after people answer them. It's extremely rude.

Answer 2

True. It is uniformly continuous on (0,1) since it is actually continuous on [0,1], which is compact. It is uniformly continuous on Z because Z is discrete, so a delta=1/2 in the definition of uniform continuity works. It is NOT uniformly continuous on R, and this is a standard example of a continuous function which is not uniformly continuous.

Answer 3

False, because the set (0, 1) and the set Z are both composed of discrete members so that if you make a change to the input to f(x) you get a jump in the resulting output. On the other hand R is a smooth set of elements, and an arbitrarily small change to the input from R into f(x) will produce an arbitrarily small change in the output. Not my area of expertise, however...

Answer 4

True.

0 is real 1 is real and anything on a parabola is either real or integer.

THERE ARE NO COMPLEX (IMAGINARY) NUMBERS!!!