4

Given $[0,1]$, is it possible to find one set $A_1\subset[0,1]$ and $A_i = A_1 + a_i, i\in\mathbb{N}, a_i\in\mathbb{R}$ such that $\{A_i\}$ is a countable partition of $[0,1]$.

By $A_1 + a_i$, it means translating $A_1$ by the value of $a_i$, e.g. $[0, 0.5] + 0.25\triangleq [0.25, 0.75]$.

Edit: It turns out with constraint of modulo one, this $A_1$ could be the non-measurable Vitali set. And $[0,1] = \cup_{q\in Q\cap[0,1]}(V_1 + q)$.

L Q
  • 424
  • I think you will find this question slightly more natural to ask about the circle. Or, equivalently, let your translation "loop around" modulo 1. At least I think it is so. – Arthur Mar 14 '22 at 18:02
  • @Arthur I find the question in its current form fascinating, honestly; not being able to work mod 1 imposes a really interesting constraint. – Steven Stadnicki Mar 14 '22 at 18:04
  • @StevenStadnicki Depends on what you want from this problem, I guess. I still find the circle more natural. – Arthur Mar 14 '22 at 18:08
  • 1
    The problem is not clear. When you translate, do you translate $\pmod 1$? Something else? But if you translate $\pmod 1$ then $0=1$ so if $A_1$ includes $0$ does it also include $1$? – lulu Mar 14 '22 at 18:20
  • @lulu Arthur, I was not thinking of modulo. But I doesn't hurt to have mod 1 for starter, in which case $[0,1]$ should be modified as $[0,1)$. – L Q Mar 14 '22 at 18:31
  • So then , the problem is standard. Just look a the equivalence relation, $x\sim y\iff x-y\in \mathbb Q$ and use the Axiom of Choice to pick a representative of each equivalence class. Then $[0,1)$ is the union of all the rational translates $\pmod 1$ of this collection. – lulu Mar 14 '22 at 18:38
  • Thiis not possible if you want to partition the closed interval $[a, b]$ : in that case, $A_1$, must contain a maximum element $t$ that is mapped onto $b$ by translation through $b - t$, but then a left neighbourhood of $b$ will only contain countably many translates of $t$ and cannot be covered by countably many translates of $A_1$. – Rob Arthan Mar 14 '22 at 19:13
  • Your edit is not quite right: if you are working modulo $1$, then $1\bmod 1$ is $0$, so $1$ is not contained in any $A_i$. You mean $[0,1)$, not $[0,1]$. – TonyK Mar 14 '22 at 19:24
  • PS: I think the original question was quite clear: if it wasn't what intended, then please edit it. – Rob Arthan Mar 14 '22 at 19:24
  • 1
    @RobArthan: I see why $A_1$ must contain a maximum element (although it is perhaps not immediately obvious). But I don't see why a left neighbourhood of $b$ will only contain countably many translates of $t$. Am I missing something? – TonyK Mar 14 '22 at 19:30
  • @TonyK - Each $A_i$ is a translate of $A_1$ and has in it only a single translate of $t$. There are only countably many $A_i$, so there are only countably many translates of $t$. But I admit that I don't see why this means only countably many translates of $A_1$ cannot cover a left neighborhood of $b$. – Paul Sinclair Mar 15 '22 at 15:36
  • @PaulSinclair: I am having trouble understanding your final sentence. It seems to contain a double negative. – TonyK Mar 15 '22 at 15:48
  • @TonyK - two negatives do not constitute a double negative when they negate different things. Rob Arthan uses the countability of the translates of $t$ to conclude that a left neighborhood of $b$ cannot be covered by countably many translates of $A_1$. I am admitting that although I see why there are only countably many translates of $t$, I do not understand why countably many translates of $A_1$ cannot cover a left neighborhood of $b$. I too am probably missing something. – Paul Sinclair Mar 15 '22 at 16:16
  • @PaulSinclair Thank you for clearing that up. – TonyK Mar 15 '22 at 17:29
  • Apologies if my comments have held up progress on this. I had doubts about my claim too, but hadn't had time to check my ideas properly until today. The line of proof I had in mind definitely does not work. In the opposite direction, I have tried to adapt Vitali's construction of a non-measurable set, but the lack of a group structure when translations are restricted to $[0, 1]$ makes that look difficult. I'd be interested to know the answer. – Rob Arthan Mar 21 '22 at 19:47

0 Answers0