Let $\mathcal{A}$ be a non-empty systems of non-empty sets such that there is an injective map $f:\bigcup \mathcal{A}\to \mathcal{A}$ such that $a\in f(a)$ for all $a\in\bigcup\mathcal{A}$. Assuming that $|\mathcal{A}| = |\bigcup \mathcal{A}|$, is there always a bijection $\varphi: \bigcup \mathcal{A}\to \mathcal{A}$ such that $a\in \varphi(a)$ for all $a\in\bigcup\mathcal{A}$. ?
$\begingroup$
$\endgroup$
3
-
1$\begingroup$ I wonder why this question got several downvotes. I upvoted. $\endgroup$– Tom De MedtsCommented Mar 18, 2015 at 8:33
-
$\begingroup$ Thanks Tom! - Apparently because there are easy examples giving a negative answer... $\endgroup$– user62017Commented Mar 18, 2015 at 11:46
-
1$\begingroup$ You may want to take a look at "Transversal Theory: An account of some aspects of combinatorial mathematics" by L. Mirsky. The whole book is about this kind of problems. Perhaps for the application you have in mind (and didn´t tell us about) there is such a $\varphi$. $\endgroup$– Ramiro de la VegaCommented Mar 18, 2015 at 17:17
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
3
No - let $\mathcal{A} = \{\{0,1\}\} \cup \{\{n\} : n\in \omega\}$.
answered Mar 18, 2015 at 6:41
-
$\begingroup$ If you mean $\{\{0, 1\}\}\cup . . .$, I don't think this works - what is the $f$? Otherwise - if we treat numbers as ordinals - 0 can't be in $\mathcal{A}$ since $\mathcal{A}$ should consist of nonempty sets only. $\endgroup$ Commented Mar 18, 2015 at 6:42
-
2$\begingroup$ I don't think this works - as I wrote in my comment, I don't think there's an injection $f$ from $\bigcup\mathcal{A}$ to $\mathcal{A}$ satisfying $a\in f(a)$, since 0 and 1 must both be sent to the same place. $\endgroup$ Commented Mar 18, 2015 at 6:59
-
$\begingroup$ Noah is right. You should erase the ",$n\geq2$" to make it work. $\endgroup$ Commented Mar 18, 2015 at 15:05
$\begingroup$
$\endgroup$
No: $A=\{\{1, 2, 3\}, \{1\}, \{2\}, \{3\}\}\sqcup \{[i, \infty)\cap \mathbb{Z}: i>4\}.$
answered Mar 18, 2015 at 6:40