g may not be continuous (recall the definition: g is continuous if g is continuous at any point of its domain). Here, g is only continuous at the point c. -1 point

You should also point out what A is. -1 point

For such type of problem, you should first say It is true, and I will prove it. Or It is not true, and I will disprove it.

-10 points since you did not prove the uniform continuity Theorem 19.2 does not apply since (0.5) is an open interval.

" |x_0-y|<delta holds" is not "because x_0 is between x and y". Instead, it is because you can choose x_0, such that x_0 satisfies |x_0-y|<delta. You can do it by picking up a rational number x_0 that is very close to y.

This is not clear due to y. If you mean for all x, and for all y, then it is the definition of uniform continuity. If you mean given a y, then for all epsilon, there exists delta and so on, it is the definition that f is continuous at y. I think you mean that f is continuous at y. So to make it clear, you should write Given y, for all epsilon>0, there exists delta>0, and so on...

Wrong. -3 points Your f(x) is not continuous on R.

Wrong. -3 points According to the Theorem 18.1, every CONTINUOUS function on a closed interval assumes its max and min values. But here, f is a discontinuous function. When the continuity condition is violated, see what will happen:

Define f(x) such that f(x)=1/2 if x is irrational. If 0<x<1, let x=p/q in lowest terms. Then let f(x)=1/q if q is even and f(x)=(q-1)/q if q is odd. Define f(1)=1/2 and f(0)=1/2.

All the irrational numbers are 0.5, so they don't matter. Otherwise the output of f is either as close as you want to 0, but never actually 0, or as close to 1 as you want, but never actually 1.

Such a function has neither an absolute minimum nor an absolute minimum on [0,1]