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Denote by $\Sigma$ the collection of all $(S, \succeq)$ wher $S \subset \mathbb{R}$ is compact and $\succeq$ is an arbitrary total order on $S$.

Does there exist a function $f: \mathbb{R} \to \mathbb{R}$ such that for all $(S, \succeq) \in \Sigma$ there exists a compact interval $I$ with the properties that

  • $f(I) = S$
  • $x \geq y$ implies $f(x) \succeq f(y)$ for all $x,y \in I$?

If so, how regular can we take $f$ to be? The motivation is that basically, I am trying to construct the analogue of a normal sequence but on $\mathbb{R}$ instead of $\mathbb{N}$.

EDIT: As Brian M. Scott points out, this is not possible if the orderings have no greatest and least elements. However, since adding this assumption doesn't go against the intuition of generalizing normal sequences, I am still interested in the answer if we restrict the various total orders to have minimal and maximal elements.

Thanks in advance.

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2 Answers 2

up vote 3 down vote accepted

If I understand correctly what you’re asking, the answer is no. Let $\preceq$ be a linear order on $[0,1]$ having no last element; no compact subset of $\Bbb R$ can be mapped onto $[0,1]$ in such a way that $f(x)\preceq f(y)$ whenever $x\le y$, because every compact subset of $\Bbb R$ has a last element with respect to the usual order. That is, if $I$ is a compact subset of $\Bbb R$, let $u=\max I$, and let $x\in[0,1]$ be such that $f(u)\prec x$; then $x\notin\operatorname{ran}f$, so $f[I]\ne[0,1]$.

Added: Restricting the linear orders to those with first and last elements doesn’t help. Well-order $[0,1]=\{y_\xi:\xi<2^\omega+1\}$. Now let $I$ be any compact interval, say $[a,b]$, and let $f:I\to[0,1]$ be an order-preserving surjection. Clearly $f(b)=y_{2^\omega}$ is the last element of in the well-ordering of $[0,1]$. Let $\langle x_n:n\in\omega\rangle$ be a strictly increasing sequence in $[a,b]$ converging to $b$. Then $\langle f(x_n):n\in\omega\rangle$ is an increasing sequence in the well-ordering of $[0,1]$, and it has a supremum, say $y_\alpha$. Then $y_\alpha<2^\omega$, since $\operatorname{cf}2^\omega>\omega$, and no $y_\beta$ such that $\alpha<\beta<2^\omega$ is in the range of $f$.

Actually, this requires a bit of modification if $f$ is not required to be strictly order-preserving (i.e., a bijection). In that case let $B=f^{-1}[\{y_{2^\omega}\}]$; $B$ must be of the form $(c,b]$ or $[c,b]$ for some $c\in I$. If $B=[c,b]$, replace $b$ by $c$ in the previous paragraph. If $B=(c,b]$, note that nothing between $f(c)$ and $f(b)$ is in the range of $f$.

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Thanks, that's a good point. I have edited this consideration into the question. –  user12014 Jun 15 '12 at 7:28

If you have the property that if $x < y$, then $f(x) \prec f(y)$, all strictly, then no.

Let $S$ be an infinite compact subset of $\mathbb{R}$. Let $\aleph$ be a cardinal in bijection with $S$. That is there is a bijection $f : \aleph \rightarrow S$. From this bijection, you can define a well-ordering on $S$ that is order-isomorphic to $\aleph$. So for our purpose, we can just think of $S$ as $\aleph$.

Assume an $f$ exists. $0 \in \aleph$ and $1 \in \aleph$. Then there exists $x < y \in \mathbb{R}$ such that $f(x) = 0$ and $f(y) = 1$. However, there are infinitely many points between $x$ and $y$ in the usual ordering on $\mathbb{R}$. However in the ordering of $\aleph$, there are no elements between $0$ and $1$. Therefore, every one of the infinitely many $z$ such that $x < z < y$, you must have that $f(z) = 0$ or $f(z) = 1$.

So if you want strict ordering to be preserved, then you can not have such a function for any possible ordering. However, with $\leq$ you can have some functions but they are not very interesting.

Moreover, you want a a single function that works for all $(S, \preceq)$ in the sense that there is some compact interval $I$ with your property. I would guess this is not possible. There are a lot of orderings of a any compact $S$. For instance given any ordering of $S$, you can make a modified linear ordering that switches the ordering of two element. This switch would require you to find a different interval $I$ that would witness your property. My intuition is that there are so many orderings and only countable many disjoint compact intervals, that there are just not enough intervals for you to define your function $f$ to make it work for all of them. But again this is not a proof just intuition.

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Thanks for pointing that out. But no, I am not so worried about strict ordering. Intuitively, I also agree that the answer should be no, but I can't prove it one way or the other. –  user12014 Jun 15 '12 at 7:30
For any set $S$, the cardinality of the total orders on $S$ is at least the cardinality of the power set of $S$ (minus a point): Pick any $s\in S$ and map an order $\preceq$ to the set $\{x\in S\colon x\preceq s\}$. This maps the set of orders onto $P(S\setminus\{s\})$. –  Harald Hanche-Olsen Jun 15 '12 at 7:34
@HaraldHanche-Olsen I am aware of the cardinality argument, but I am not sure if a cardinality argument alone suffice to prove that the function does not exists. I am not sure if overlapping compact intervals $I$ may somehow work as a witness for different $(S, \preceq)$. –  William Jun 15 '12 at 7:37
Sorry, I wasn't sure myself exactly what you were unsure about in your final paragraph. –  Harald Hanche-Olsen Jun 15 '12 at 12:28

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