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Assume AC. Let $x_\alpha$ be a well-ordering of $\mathbb{R}$. For all $\alpha < \mathfrak{c}$, let $F(x_\alpha) = x_{\alpha+1}$.

Can it be proven that $F$ is discontinuous everywhere?

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    Also, can it be shown that $F(x_\alpha) = x_{\alpha+1}$ is _equivalent_ to saying $F$ has no fixed points?2012-11-03

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$F$ can be continuous at some points, if the well-ordering is defined in the right way. For example, choose your well ordering so that each real number in $(0,1)$ is a unique limit ordinal in $c$. That is, if $T:\mathbb{R} \rightarrow c$ maps each element $x \in \mathbb{R}$ to the element $\alpha$ of $c$ such that $x_\alpha=x$, then we want $x \in (0,1) \Rightarrow \alpha$ is a limit ordinal or 0.

There are enough limit ordinals to accomplish this, because the cardinality of $c$ is the cardinality of $\mathbb{N} \times L(c)$ where $L(c)$ is (the set of all limit ordinals in $c$) $\cup$ 0. So the number of limit ordinals must have the same cardinality as $c$, and the same cardinality as (0,1).

Further define $T$ so that $T(x) = T(x-2)+1$ for all $x \in (2,3)$. At this point $T$ is still injective, because $T(x-2)+1$ is not a limit ordinal.

Now extend $T$ to the rest of $\mathbb{R}$ where it hasn't already been defined in such a way that it is bijective.

With $x_\alpha$ defined in terms of this $T$, F will be continuous on $(0,1)$. In fact, it will be identically equal to $f(x)=x+2$ on $(0,1)$.

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    you can do it for $R^+$, but not for R2012-11-04