Recent content by splash_lover

but I need to get rid of the f'(x) and the (x-x0). How do I do that?

1. Prove f is differentiable at x=xo implies f is continuous at x=xo using epsilon and delta notation. 2. I have gotten this far: absolute value(f(x)-f(xo)) <= absolute value(x-xo)*(epsilon + absolute value(f '(xo))) <= means less than or equal to. 3. I need to get here: absolute...

No I don't see how I can use it here in this problem. How would I start my step two? I know I assume alpha belongs to S closure, but I am not sure where to go from there.

Given that alpha is an upper bound of a given set S of real numbers, prove that the following two conditions are equivalent: a) We have alpha=sup(S) b) We have alpha belongs to S closure I'm trying to prove this using two steps. Step one being: assume a is true, then prove b is true...

no, because some of the points in U (set of interior points)are not included in the original set S.

Is it even possible to find an example for part c? I know the example I gave is wrong.

So the closure is [0,2]. Was the example i gave for part C correct?

no, (0,2) can't be the interior of S. So it would be (0,1)U(1,2)?

no it can't because 1 is not included in S

I think so, when I read the problem that's all it had was S=[0,1)U(1,2). So I am assuming (1/2,3/2) is contained in S.

Suppose that S=[0,1)U(1,2) a) What is the set of interior points of S? I thought it was (0,2) b) Given that U is the set of interior points of S, evaluate U closure. I thought that U closure=[0,2] c) Give an example of a set S of real numbers such that if U is the set of...