If the range is {-1,1}, you can find x,y in [a,b] such that f(x)=-1 and f(y)=1. f can't be continuous in the interval [x,y] (or [y,x]!) because of Bolzano.
I need to show that if f is a function with domain on the closed interval [a,b] and the range of f is {-1,1}, then there is a number x in [a,b] at which f is not continuous.
I'm guessing it is not continuous at a or b or both but I'm not sure how to go about proving it.
In the class I am in I am unable to reference the IVT, so I need to prove it some other way.
Oh and another question. if f(a)=-1 and f(b)=1 why can't this be continuous? it could just be a straight line from (a,-1) to (b,1) and not jump around.....right?
You are misreading the problem. You said "I guess I don't see why f taking on all values between -1 and 1 on [x,y] means it's discontinuous if the range is from -1 to 1."
The range is NOT "from -1 to 1". That would be [-1,1]. The problem said that the range is {-1, 1} which is NOT an interval- it is the two numbers, -1 and 1, only.
Well, then, I guess the problem depends upon what definition of "continuous" you are using. One regularly useddefinition is "f":A->B is open if and only if for every open subset, U, of B, is an open subset of A". Here, the topology for B= {-1, 1}, since it contains of finite number of points, is the "discrete topology". Every subset is open which means that, if f is continuous, every subset of [-1, 1] must be continuous. That is not true.
The definition of continuity we are using is: The function f is continuous at the point p=(x,y) if p is a point on f and if epsilon is any positive number, then there is a positive number delta such that if t is in the domain of f and the absolute value of t-x is less than delta, then the absolute value of f(t)-f(x) is less than epsilon.
Redsox fan kind of used this one, but I was told by my prof that this assumes uniform continuity (?) and I'm not allowed to do so.
Again, thanks for responses.