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We say that a function $f: \mathbb{R} \rightarrow \mathbb{R}$ preserves the binary relation $\sim \subseteq \mathbb{R}^2$ if $x \sim y$ implies $f(x) \sim f(y)$ for all $x,y\in\mathbb{R}$.

We say that $\sim$ ensures continuity if every function $f: \mathbb{R} \rightarrow \mathbb{R}$ that preserves $\sim$ is continuous.

Part 1. Is there any binary relation that ensures continuity?

A well-ordering would work for this, I think (edit: Silly me, I realized this is not actually true. But a totally rigid relation exists on any set given Choice anyway.) But I'd like to see something better (in Part 2), and if possible something that doesn't require Choice for Part 1.

Part 2. Is there any binary relation that ensures continuity, and is preserved by at least two different functions?

Z. A. K.
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    An explanation for the downvote would be much appreciated. Am I missing some very simple construction here? – Z. A. K. Sep 17 '23 at 15:41
  • Your point about the well-ordering is that it is only preserved by the identity function, right? – Alex Kruckman Sep 17 '23 at 15:44
  • @AlexKruckman: That's right (I also had what I thought was a choice-free argument for creating a totally rigid digraph there, but I then realized that it used an assumption that required choice, so I deleted it. Unfortunately, that deleted part contained the observation about the identity function.) – Z. A. K. Sep 17 '23 at 15:45
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    There are some interesting examples if you allow relations of higher arity. For instance, a function preserves the ternary relation $x<y<z$ and $z-y=y-x$ iff it is linear with positive slope. – Eric Wofsey Sep 17 '23 at 17:58
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    Similar to Eric's example you can also do it with infinitely many relations, e.g. if $f$ has to preserve the relations $|x - y| = d$ for all $d \ge 0$ then $f$ is an isometry and so has the form $f(x) = \pm x + a$. – Qiaochu Yuan Sep 17 '23 at 21:20
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    @EricWofsey: Indeed, many ternary relations work. In turn, one can show easily that no unary relation could work. It's why I' got so excited about what happens in the binary case :) – Z. A. K. Sep 18 '23 at 02:01
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    It might be interesting to note that the Beckman-Quarles Theorem tells you that the binary relation of being unit distance from each other being preserved is equivalent to $f$ being an isometry when working in $\mathbb{R}^n$ for $n>1$ (but this condition obviously is not sufficient even for continuity over $\mathbb{R}$). – Isky Mathews Sep 21 '23 at 21:47
  • The first obvious guess would be to consider the usual ordering relation $\le$. But in fact, there are functions that are monotonic but not continuous. – Geoffrey Trang Sep 24 '23 at 17:57
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    Not sure if this is allowed but what a fantastic question! – Carlyle Sep 26 '23 at 20:07

1 Answers1

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This answer was simplified by a lot from the original one.

Consider a relation defined as follows: $x \sim y$ if $x > y$, or $y = x + 2^{m}$ for some $m\in\mathbb{Z}$.

The idea is that the first part of the definition will force the function to be strictly increasing, the second part constrains what the values of $f(x+2^m)$ might be in terms of $f(x)$. This then gives us a single sequence along which $f(x+2^m)$ converges to $f(x)$ and monotonicity then forces $f$ to be sequentially-continuous.

First, let's demonstrate that any function preserving this relation is strictly increasing. Assume $x > y$, if $f(x) \leq f(y)$ then we must have $f(y) = f(x) + 2^{m}$.

Pick $z$ such that $x>z>y$, then we have $f(x) \sim f(z) \sim f(y)$. There are uncountably many such $z$ so there must be uncountably many $z$ satisfying $f(x) > f(z)$, since otherwise we would have $f(z) = f(x) + 2^{m}$ for some $m(z)$. This implies $m(z_1) = m(z_2)$ for uncountably many pairs $z_1, z_2$. Since $f(z_1) \sim f(z_2)$ as well as $f(z_1) = f(z_2)$, we have a contradiction, as $a\sim a$ never holds for our relation.

We apply this same argument to all $z$ satisfying $f(x) > f(z)$, to establish $f(z) > f(y)$ for some $z$. Therefore, we have successfully shown $f(x) > f(y)$.

Next, let's demonstrate that any function preserving this relation is continuous. Since $x \sim x + 2^{m}$, we have $f(x) \sim f\left(x+2^{m}\right)$. As $f$ is strictly increasing, this can only occur if $f\left(x+2^{m}\right) = f(x) + 2^{k(m)}$. Therefore, $\lim_{m\to-\infty} f\left(x+2^{m}\right) = f(x)$, as $k(m)$ is strictly increasing. Consequently, $f$ being strictly increasing gives us right continuity.

Similarly, since $x - 2^{m}\sim x$, we have $f\left(x - 2^{m}\right) \sim f(x)$. Again, considering $f$ is strictly increasing, this can only occur if $f\left(x - 2^{m}\right) = f(x) - 2^{k(m)}$. Thus, $\lim_{m\to-\infty} f\left(x - 2^{m}\right) = f(x)$, since again $k(m)$ is strictly increasing. This leads to left continuity.

So we have shown that $f$ is continuous. Furthermore, $f(x) = 2^n x$ for any $n$ preserves this relation.

Jan Kuś
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  • Thank you so much for this answer. It came in a really busy time: I'll make some time to read and understand it asap, in a a day or two. (Admin question @Noah: in case you decide in the meantime to award bounty for this, can you do if the answer is not/not-yet accepted?) – Z. A. K. Sep 24 '23 at 06:34
  • +1. You can extend this to include all linear functions with rational gradients, or in fact gradients from any countable subset of $\mathbb{R^+}$ that is closed under multiplication, and has at least one element less than $1$ (guaranteeing that the set has $0$ as a limit point) – Carlyle Sep 27 '23 at 12:31
  • @Carlyle I think this only works when $0$ is the only limit point of your sequence. Otherwise say $q_n$ is the enumeration of your set and it also has (say) $1$ as it's limit point.Then $f(x+q_n)=f(x)+q_{k(n)}$. But $q_k(n)$ could then converge to $1$, so that $\lim_{n\to\infty} f(x+q_n)=f(x)+1$. – Jan Kuś Sep 27 '23 at 15:53
  • @Jank493 yes, that is why I said it should contain an element $x$ which is less than $1$, combining this with closure under multiplication means $x^k$ is also in the set, which gets arbitrarily small when $x<1$. – Carlyle Sep 27 '23 at 16:15
  • @Carlyle I still don't see how you get continuity for such sets. Take positive rationals as your set which are closed under mulitplication, then how do you rule out the existance of functions for which $f(x+2^{-n})= f(x)+(\frac{1}{2}+\frac{1}{n})$ or similar identities holding? These would be consistent with my solution, unless the only limit point is $0$. – Jan Kuś Sep 27 '23 at 19:41
  • Oh I see what you mean, you are right it doesn't work. – Carlyle Sep 27 '23 at 19:44
  • @Jank493: No pressure, but have you considered using your full name in your profile? This is a nice example that people might want to attribute to you. – Z. A. K. Sep 30 '23 at 01:26