Proving Continuous Functions Cannot Be Two-to-One

[jeff1evesque]

Homework Statement

Suppose $$f: [0,1] \rightarrow R$$ is two-to-one. That is, for each $$y \in R, f^{-1}({y})$$ is empty or contains exactly two points. Prove that no such function can be continuous.

Homework Equations

Definition of a continuous function:
Suppose $$E \subset R$$ and f: E \rightarrow R[/tex]. If $$x_0 \in E$$ then f is continuous at $$x_0$$ iff for each $$\epsilon > 0$$, there is $$\varphi >0$$ such that if

$$|x - x_0| < \varphi, x \in E$$,​

then

$$|f(x) - f(x_0)| < \epsilon$$.​

If f is continuous at x for every $$x \in E$$, then we say f is continuous.

The Attempt at a Solution

I think from an algebraic view, there is more elements in the domain then codomain which means that the function f is not one-to-one, but onto. Though I know this fact, I am pretty sure this will not aid in my attempt to prove this problem. Could someone help me understand why f cannot be continuous.


Thanks,


Jeffrey Levesque

 

[foxjwill]

Try assuming that f is continuous and then using the extreme and intermediate value theorems to produce a y in R with more than two values in its pre-image.

 

[HallsofIvy]

Prove, rather, that there exist y such that [math]f^{-1}(\left{y\right})[/math] contains exactly one value.