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Let $\varepsilon>0$ be a positive number. My goal is to prove the following: there does not exist a finite collection of functions $\{f_i:\mathbb R\to\mathbb R\mid i\in\{1,\ldots,n\}\}$, where $n\in\mathbb N$, such that the following condition holds: for any $x,y\in\mathbb R$,

$$x>y+\varepsilon\quad\text{ if and only if }\quad f_i(x)\geq f_i(y)\text{ for all $i$, with strict inequality $>$ for some $i$}.$$

That is, the presence of “gaps” of width greater than $\varepsilon$ between two numbers cannot be represented, in a sense, by any finite family of functions.

Any hints or comments would be appreciated.

UPDATE #1 Here is what I could prove so far:

Suppose such a finite family of functions exists for some $n\in\mathbb N$. Then, for each $x\in\mathbb R$ and $\delta\in(0,\varepsilon]$, there is some $i\in\{1,\ldots,n\}$ (depending on $x$ and $\delta$) such that $$f_i(x)>f_i(x+\delta).$$

My conjecture is that the desired contradiction would build on there being only finitely many functions in the family, so that there would exist some $i$ for which the inequality above held for infinitely many $x$ and $\delta$, but I cannot see exactly how this leads to a contradiction.

UPDATE #2 I managed to further reduce the problem to the following conjecture:

Fix $\varepsilon>0$. There does not exist a function $f:\mathbb R\to\mathbb R$ such that the set $$\{x\in\mathbb R\mid f(x+\varepsilon+\xi)\geq f(x)>f(x+\varepsilon)\text{ for all $\xi>0$}\}$$ is uncountable.

triple_sec
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  • Can you solve the case $n=2$? – Noah Schweber Nov 23 '17 at 03:56
  • @NoahSchweber I think so, based on the accepted answer to this question. However, I’m not sure how it generalizes to $n\geq 3$. Indeed, the problem I linked does not generalize (given that there exists a family of three functions satisfying the claim), but the sufficient condition in that problem is more stringent. I still think non-existence in the current problem holds, but I’m stuck with the general finite case. If you have any ideas, I’d more than welcome them. – triple_sec Nov 23 '17 at 04:02

1 Answers1

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I got it.

I am first going to prove that the conjecture put forth in the second update is true. Take any function $f:\mathbb R\to\mathbb R$ and define $$E_f\equiv\{x\in\mathbb R\mid f(x+\varepsilon+\xi)\geq f(x)>f(x+\varepsilon)\text{ for all $\xi>0$}\}.$$ For each $x\in E_f$, let $q_x$ be a rational number satisfying $f(x+\varepsilon)<q_x<f(x)$. Next, suppose that $x,y\in E_f$ and $y>x$. Then, one has that $$f(y)>q_y>f(y+\varepsilon)=f(x+\varepsilon+\underbrace{y-x}_{>0})\geq f(x)>q_x>f(x+\varepsilon).$$ This shows that the mapping $x\mapsto q_x$ from $E_f$ to $\mathbb Q$ is injective, so that $E_f$ is countable.

Actually, this can be used to prove that there does not exist even a countable family $\{f_i:\mathbb R\to\mathbb R\}_{i\in\mathbb N}$ such that $$x>y+\varepsilon\quad\text{ if and only if }\quad f_i(x)\geq f_i(y)\text{ for all $i$, with strict inequality $>$ for some $i$}.\tag{$\spadesuit$}$$ If such a family existed, then, for each $x\in\mathbb R$, there were some $i\in\mathbb N$ such that $f_i(x)>f_i(x+\varepsilon)$. If not, it would be the case that $f_i(x+\varepsilon)\geq f_i(x)$ for all $i\in\mathbb N$. But since $x+\varepsilon$ is not greater than itself, ($\spadesuit$) yields that $f_i(x+\varepsilon)= f_i(x)$ for each $i\in\mathbb N$. But then $(\spadesuit)$ again implies that $$f_i(x+2\varepsilon)\geq f_i(x)=f_i(x+\varepsilon)$$ for all $i\in\mathbb N$, with strict inequality for some $i$. This, in turn, implies, again by ($\spadesuit$), that $x+2\varepsilon>(x+\varepsilon)+\varepsilon$, which is clearly impossible. What I just proved is that $$\bigcup_{i\in\mathbb N} E_{f_i}=\mathbb R,$$ which is at odds with $E_{f_i}$ being countable for each $i\in\mathbb N$.

triple_sec
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