I ask this after trying to solve it myself unsuccessfully. It's part of a larger question I'm working on, but have separated this out so as to make it as clear and concise as possible.

Trying to solve this: $$\text{Let }r \in \, (0,1\, ) \text{ and } r \notin \mathbb{Q}$$ $$\text{Let }r’ \in \, (0,1\, ), r’>r, r’ \notin \mathbb{Q}, \text{ and } r’-r \notin \mathbb{Q}$$ $$\text{Let }q \in A = \{ r’-r+p : p \in \, (r,r+1\, ) \cap \mathbb{Q} \}$$ Finally, one of the two statements must be true: $$\text{1) } \exists q’ \in B = \, (r’,r’+1\, ) \cap \mathbb{Q} \text{ such that }\{q'-p : p \in B\} = \{q-p : p \in A\}$$ $$\text{2) } \nexists q’ \in B = \, (r’,r’+1\, ) \cap \mathbb{Q} \text{ such that }\{q'-p : p \in B\} = \{q-p : p \in A\}$$

Which statement is true?

  • $\begingroup$ I suggest you start by picking specific values of $r,r',q...$ etc. that satisfy the above conditions, and try to understand the problem that way, drawing a picture as you go. It is difficult to keep straight in your head what all the different variables are. $\endgroup$ – Squeeze my theorem Mar 25 '18 at 16:03
  • $\begingroup$ I have: specific values, pictures, and everything. I've attempted to assume statement 1 is true so as to try to arrive at a contradiction, but don't. I've also attempted to assume statement 2 is true so as to try to arrive a contradiction, but don't. It seems undecidable as far as I can tell. I don't think it's like defining a flying unicorn and asking whether it exists though. I think it's a valid question because everything is well defined. Any help is appreciated. $\endgroup$ – AplanisTophet Mar 25 '18 at 16:21

The second statement is true.

Let $q'\in B$. Then $$\inf(\{p-q': p\in B\})-\inf(\{p-q:p\in A\})=(\inf B-q')-(\inf A-q)=(r'-q')-(r'-q)=q-q'\neq 0, $$ because $q'\in\mathbb{Q}$ and $q\notin\mathbb{Q}$. So, the sets are different as they have different infima.

Edit: The answer to the corrected question is the same, as $$\sup(\{q'-p:p\in B\})-\sup(\{q-p:p\in A\})=(q'-\inf B)-(q-\inf A)=q'-q. $$

  • $\begingroup$ You gave me exactly what I was looking for. This solves the other question I am working on too! Showing that the difference between the infima, and that the difference must be a real number (in this case an irrational number) is exactly what I was going for. $\endgroup$ – AplanisTophet Mar 25 '18 at 20:56

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