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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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    Is $F$ intended to be defined on $x_\alpha$ where $\alpha \geq c$? If so, how is it defined?2012-11-02
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    I guess I'm confused. Unless I misunderstand, as phrased there is no guarantee that $F$ is defined on all of $R$. Equipollent is a cardinal relation, so even though $c$ is equipollent to $R$ it might not be order-isomorphic to the well ordering on $R$. There are many ordinals equipollent to $R$ and by choosing the smallest you make it likely that many elements of $R$ won't be included in the definition of $F$.2012-11-02
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    $x_\alpha$ is a bijection between $c$ and $\mathbb{R}$, constructed via the well-ordering principle, so $F$ is also defined on all of $\mathbb{R}$ and is a bijection itself. Maybe it is easier to discuss if we just assume CH, as long as that doesn't influence the answer.2012-11-02
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    There are many well orderings on $\mathbb{R}$, not all of them are bijections between $c$ and $\mathbb{R}$. If you want your well ordering to have that property might want to edit your question to include it.2012-11-02
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    What I mean by a well-ordering on $\mathbb{R}$ is a bijection between the minimal ordinal of $\mathbb{R}$ and $\mathbb{R}$. AC implies that such a function exists, I am denoting it as the transfinite sequence $x_\alpha$.2012-11-02
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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/6308/discussion-between-dan-brumleve-and-logan)2012-11-02
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    Dan, I feel that the answer is yes or it depends on CH in some way. I have no idea what the argument should be, though. Interesting question, by the way (+1).2012-11-02
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    For some reason noone has pointed this out yet: $F$ is not a bijection. Its range misses $x_0$ and all $x_\alpha$ with $\alpha$ limit ordinal.2012-11-02
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    @Stefan, you are right and I was mistaken about that! But I think the question is still valid as stated.2012-11-02
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    @Dan Brumleve: Yes, sure, the question is completely valid. I just wanted to clear up this misconception.2012-11-02
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    $F(x_\alpha)=x_{\alpha+1}$ implies that $F$ has no fixed points. Would it be possible to use some fixed-point theorem to show that it isn't _continuous_ everywhere? Would any additional conditions be required? My interpretation of @Logan's answer yields a function which remains discontinuous at one point.2012-11-03
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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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    Points whose index is not a successor ordinals are not in the image of $F$. So if you make $(0,1)$ into limit ordinals you just ensured $F(x)\notin(0,1)$ for all $x\in\mathbb R$.2012-11-02
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    @AsafKaragila I wasn't trying to put $F(x) \in (0,1)$. I was trying to get $F$ to map $(0,1)$ onto (2,3) continuously. $T$ is a bijection sending (0,1) to limits and (2,3) to successors of limits, and $F$ is a continuous function R->R sending (0,1) to (2,3).2012-11-02
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    I think this works fine for the domain $\mathbb{R}^+$. That is good enough for me, but I am going to work through it a little more before accepting.2012-11-03
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    The domain can be $\mathbb{R}$ if we map $[-1,1)$ to $L(\frak c)$ instead, counting down from $[-1,0)$ and counting up from $[0, 1)$. In that case $f(x) = x+1$ when $x \ge 0$ and $f(x) = x-1$ when $x \lt 0$ and it has a single discontinuity at $x = 0$.2012-11-03
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    as described here, the domain can be literally anything that contains (0,1)$\cup$(2,3) and has the cardinality of the continuum.2012-11-03
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    @Logan, your crucial insight as I see it is that an interval of $\mathbb{R}$ can be bijected with the limit ordinals in $\frak c$ just as easily as with $\frak c$ itself, and then the rest of $\mathbb{R}$ can be covered with translations of this interval that biject with the successor ordinals. But I don't quite understand your specific construction. For example, it doesn't seem like T($\frac{3}{2}$) is defined no matter how I read it. Should the second paragraph be understood to define $T$ on $(0,1)$? Why aren't the intervals half-open so the whole line can be covered?2012-11-03
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    T(3/2) is defined in the 4th paragraph. I don't specifically say how to define it, it doesn't matter. I constructed a pretty minimal counterexample. You are right that a better counterexample could easily be constructed, but half-open intervals wouldn't help because no matter what the function will be discontinuous at some of the integers. You can define T so that [0,1) is sent to the limit points, and then map successors to integral translations, so that F is continuous at all non-integers, but to get continuity at say 0 you need [-1,0) and [0,1) sent to [n-1,n) and [n,n+1), which is hard2012-11-04
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    you can do it for $R^+$, but not for R2012-11-04