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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? –  Logan Stokols Nov 2 '12 at 6:39
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 $F$. –  Logan Stokols Nov 2 '12 at 6:46
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). –  Asaf Karagila Nov 2 '12 at 7:16
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. –  Stefan Geschke Nov 2 '12 at 7:43
@Dan Brumleve: Yes, sure, the question is completely valid. I just wanted to clear up this misconception. –  Stefan Geschke Nov 2 '12 at 7:49

1 Answer 1

up vote 2 down vote accepted

$F$ 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 $x\in\mathbb R$. –  Asaf Karagila Nov 2 '12 at 7:57
@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). –  Logan Stokols Nov 2 '12 at 8:05
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. –  Dan Brumleve Nov 3 '12 at 17:01
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 $x = 0$. –  Dan Brumleve Nov 3 '12 at 18:16
as described here, the domain can be literally anything that contains (0,1)$\cup$(2,3) and has the cardinality of the continuum. –  Logan Stokols Nov 3 '12 at 20:35

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