0
$\begingroup$

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 $\cup_L L_n$

  • 0
    What about $\forall L_i \in L: x \notin L_i$? Is that what you mean? – 2017-01-24
  • 0
    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? – 2017-01-24
  • 1
    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. ^^ – 2017-01-24
  • 1
    $\cup_{n\in\mathbb{N}}\{L_n\}$ 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. – 2017-01-24

1 Answers 1

1

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 $x$: $$\forall \Lambda\in L:x \notin \Lambda\\ \neg\exists\Lambda \in L:x \in \Lambda$$

  • 0
    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! – 2017-01-24
  • 1
    Maybe a typo but shouldn't it be $\neg\exists\Lambda \in L:x \in \Lambda$ ? – 2017-01-24
  • 1
    I would say $$x\notin \bigcup_{L_n\in L}L_n$$ is more natural and readable. – 2017-01-24
  • 0
    @TonyK I agree, that's what I'm going with. – 2017-01-24