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Provide an example of a non-empty transitive relation $r$ on the set $\mathbb N$ such that the relation $r^{\exists}$ defined in the set $\mathcal{P}$($\mathbb N$) by the condition: $$\langle X, Y \rangle \in r^{\exists} \iff \exists x \exists y (x \in X \; \wedge \; y \in Y \; \wedge \langle x, y\rangle \in r)$$ (a) is transitive; (b) is not transitive.

I started with definition of transitive relation. As I understand, $r^{\exists}$ should have 2 conditions: it should be transitive and have condition above (am I right?). But I don't understand how to connect $r$ with $r^{\exists}$?

Any help will be much appreciated.

  • Use *text* to produce italics, not MathJax. – Arturo Magidin Nov 28 '23 at 19:54
  • Have you tried some examples (forget about "transitive" and "non-transitive", just some examples...)? For instance, if $r$ is the relation $x\leq y$, what is $r^{\exists}$? – Arturo Magidin Nov 28 '23 at 19:56
  • PS Something is missing. You have "so that the relation $r^{\exists}$...." but you don't say what is supposed to happen to the relation $r^{\exists}$. – Arturo Magidin Nov 28 '23 at 20:22
  • @ArturoMagidin Relation $r^{\exists}$ is just determined by condition. There is no anything about this "what". –  Nov 28 '23 at 20:30
  • That is terrible phrasing, then. If I say "Find examples of $r$ so that $r^{\exists}$..." then I expect some property that $r^{\exists}$ is expected to have. If all you have is "given an example of a transitive and a nontransitive relation and then compute the corresponding $r^{\exists}$", then whoever wrote the instructions needs to edit them. If that is the case, then your understanding is incorrect: the property of being transitive and nonstransitive is what $r$ must satisfy, not what $r^{\exists}$ must satisfy. What you have above is not "condition" on $r^{\exists}$, it's a definition. – Arturo Magidin Nov 28 '23 at 20:37
  • OK. So I should find $r$ that is transitive and prove that exists at least one $r^{\exists}$ determined by condition? –  Nov 28 '23 at 20:40
  • That's nonsensical. Given a relation $r$ on $\mathbb{N}$, $r^{\exists}$ necessarily exists, defined by the condition you give. The instructions appear to be to take a nonempty transitive relation $r$, and describe the corresponding $r^{\exists}$; and to then take a non-transitive relation $r$, and describe its corresponding $r^{\exists}$. This is a very badly phrased and as far as I can tell rather dumb problem. That is why I think what you have quoted is either misquoted or not what was intended. – Arturo Magidin Nov 28 '23 at 20:43
  • I have translated it from another language and maybe because of that it sounds a bit weird. But in this "another language" it sounds even worse imo. I will try to retranslate this and add in question. –  Nov 28 '23 at 20:48
  • What language? Many people here are multilingual. Post the original in addition to your translation. – Arturo Magidin Nov 28 '23 at 20:49
  • I have added translation as it should be. And also I think should we move this discussion to chat? P. S Original language is polish –  Nov 28 '23 at 20:56
  • See, you did have a condition that needed to apply to $r^{\exists}$. This phrasing actually makes sense. So your task is: start with a transitive relation $r$ that is not empty. That relation will completely and fully determine the relation $r^{\exists}$. That relation, $r^{\exists}$ may be transitive or not transitive. You want to find an example where $r^{\exists}$ is transitive, and an example where $r^{\exists}$ is not transitive. You should start by doing some examples starting with an $r$ you know and figuring out what $r^{\exists}$ will be then, to get a feel for it. – Arturo Magidin Nov 28 '23 at 21:01

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