So if in the Collatz conjecture $L=\{L_1,\ldots\}$ is the set of all sets of elements of loops and $L_1=\{1,2,4\}$ but we do not know if there exist $L_2,\ldots$ then how do I write that $x$ is not an element of a loop?

Clearly $x\notin L$ is wrong. I guess I need to write $x$ is not in the union of elements of $L$.

Might this be $\cup_{n\in\mathbb{N}}\{L_n\}$ perhaps?

Or $\cup_{L_n\in L}L_n$

Or simply $\cup_L L_n$

  • $\begingroup$ What about $\forall L_i \in L: x \notin L_i$? Is that what you mean? $\endgroup$ – johnnycrab Jan 24 '17 at 12:19
  • $\begingroup$ Yes that would do it. Would that be the normal way, or is there also a way to notate the union over a family of sets? $\endgroup$ – samerivertwice Jan 24 '17 at 12:20
  • 1
    $\begingroup$ I don't know if this is the "normal" way. Your notation $x \notin \cup_{L_n\in L}L_n$ is not wrong either, of course. I would depend it on the flow of reading. ^^ $\endgroup$ – johnnycrab Jan 24 '17 at 12:25
  • 1
    $\begingroup$ $\cup_{n\in\mathbb{N}}\{L_n\}$ is just $L$. It should be $\cup_{n\in\mathbb{N}}L_n$, without the curly brackets. Apart from this, all your suggestions are equally good, I would say. $\endgroup$ – TonyK Jan 24 '17 at 12:29

So you want to say that $x$ is not in the union of all members of $L$?

$$x~\notin~ \bigcup_{\Lambda\in L}\Lambda$$

This is equivalent to asserting that there is no set in $L$ which contains $x$: $$\forall \Lambda\in L:x \notin \Lambda\\ \neg\exists\Lambda \in L:x \in \Lambda$$

  • $\begingroup$ Ok thanks so that's pretty much what I had but you've used a big $\cup$ instead of a little one. But now it looks much better! $\endgroup$ – samerivertwice Jan 24 '17 at 12:25
  • 1
    $\begingroup$ Maybe a typo but shouldn't it be $\neg\exists\Lambda \in L:x \in \Lambda$ ? $\endgroup$ – Zubzub Jan 24 '17 at 12:31
  • 1
    $\begingroup$ I would say $$x\notin \bigcup_{L_n\in L}L_n$$ is more natural and readable. $\endgroup$ – TonyK Jan 24 '17 at 12:31
  • $\begingroup$ @TonyK I agree, that's what I'm going with. $\endgroup$ – samerivertwice Jan 24 '17 at 13:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.