Proving Two-to-One Function Not Continuous

[dmuthuk]

Hi, I've been trying to prove this statement for a while now but haven't made much progress:

Suppose f:[a,b]\to\mathbb{R} has the property that for each y in the image of f, there are EXACTLY two distinct points x_1,x_2\in [a,b] that map to it. Then, f is not continuous.

Well, I considered two approaches. I figure it has to do with the intermediate value theorem somehow, but I could be wrong. The argument that I'm hoping would do the trick goes as follows and it is a proof by contradiction approach:

We shall assume that f is continuous. Construct a function g:[a,b]\to[a,b] as follows. Each x\in [a,b] has a unique partner x'\in[a,b] such that f(x)=f(x') via the two-to-one property of f. So, we can have g:x\mapsto x' and g is also bijective. Now, here's the problem. I have an instict that g is continuous because of f but I don't know how to prove it. If this is indeed true, then we're done because of the fixed point theorem.

Any suggestions would be appreciated, thanks:)

 

[mathman]

Suggestion: [a,b] can be divided into two intervals L and U as follows. For each pair (x1,x2) that have the same image and x1 < x2, put x1 in L and x2 in U. I suspect all points in L are then < all points in U (you will need to use the fact that there are exactly two points going into the image). Since they must make up the entire interval [a,b], L and U will meet at some point x. There will be no other point x' such that f(x)=f(x').