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I want to find a function $f:[0,1] \to [0,1]$ such that $f$ takes on each value in $[0,1]$ exactly twice. I think this means there are an infinite number of discontinuities. Can anyone help me figure this one out?

Anyone have any pointers?

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    @coffeemath: Nice!2012-11-02

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Let $x_\alpha$ be a well-ordering of $[0,1]$.

For any ordinal $\alpha = \theta + n < \frak{c}$ where $\theta$ is a limit ordinal or $0$ and $n$ is a finite ordinal, let $F(\theta + n \cdot 2) = F(\theta + n \cdot 2 + 1) = x_\alpha$.

Now define $f(x_\alpha) = F(\alpha)$ for all $\alpha \lt \frak{c}$ and it is clear that $f$ has the required property.

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    This is a overblown considering @EuYu's comment. I do need practice with this kind of proof but it's not appropriate for this problem.2012-11-02