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jeff1evesque

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## Homework Statement

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

## Homework Equations

Definition of a continuous function:

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

[tex]|x - x_0| < \varphi, x \in E[/tex],

then[tex]|f(x) - f(x_0)| < \epsilon[/tex].

If f is continuous at x for every [tex]x \in E[/tex], 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

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