Continuous functions

jeff1evesque

1. The problem statement, all variables and given/known data
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.

2. Relevant 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.

3. 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

Last edited:
Related Calculus and Beyond Homework News on Phys.org

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 $f^{-1}(\left{y\right})$ contains exactly one value.

"Continuous functions"

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving