1. Continuity

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.

2. 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.

3. I would have gone with IVT. Assume $f$ is continuous. If there exists $x,y\in[a,b]$ such that $f(x)=1$ and $f(y)=-1$, then $f$ takes on all real values between $1$ and $-1$ on the interval $[x,y]$. But this is impossible since the range is only $\{-1,1\}$. So $f$ is discontinuous.

4. redsox...could you elaborate on your response? specifically, this part... "then takes on all real values.... But this is impossible since the range is only ".

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.

5. Originally Posted by spikedpunch
redsox...could you elaborate on your response? specifically, this part... "then takes on all real values.... But this is impossible since the range is only ".

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.
It is just this simple.
How can a function be continuous if it ‘jump’ back and forth between $-1~\&~1$.
That violates the intermediate value theorem.

6. 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?

7. Originally Posted by spikedpunch
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?
Set $\epsilon=1$. $\forall\delta>0$, $\exists x,y\in[a,b]$ such that $f(x)=1$, $f(y)=-1$, and $|x-y|<\delta$ (since the range of $f$ is only the two points $\{-1,1\}$, NOT the interval of values $[-1,1]$).

Thus $|x-y|<\delta$ and $|f(x)-f(y)|=2>1=\epsilon$, so $f$ is discontinuous.

8. I looked into your answer a bit and apparently it assumes uniform continuity, which hasn't been proven in my class and therefore cannot be used. The only tool I have to show continuity is the epsilon-delta way at a point x.

So I'm back to square one.

9. 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.

10. I understand that, thanks. But that doesn't really help or answer my question. The only definition I can use to prove discontinuity is epsilon-delta, and I can't use uniform continuity or the IVM.

Thanks to any and all responses

11. 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, $f^{-1}(U)$ 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.

12. 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.