2
$\begingroup$

Let $\mathbb D$ and $\mathbb E$ be two directed sets, then a map $f:\mathbb D\to \mathbb E$ is said to be final,if for any $e\in E$ there exists some $d\in D$ such that $f(d')\geq e$ whenever $d'\geq d.$

Question: For two arbitrary directed sets $\mathbb D$ and $\mathbb E,$ does there exists a third directed set $\mathbb A$ such that there exists two final maps $f_1:\mathbb A\to \mathbb D$ and $f_2:\mathbb A\to \mathbb E?$ Any help would be appreciated.

$\endgroup$
  • 1
    $\begingroup$ It seems like this third directed set would act like the categorical product of the first two.. is that true ;-) $\endgroup$ – Musa Al-hassy Dec 24 '18 at 15:50
1
$\begingroup$

Let $\mathbb{A}$ be the family of finite subsets of $\mathbb{D}\cup \mathbb{E}$ ordered by inclusion. This is clearly a directed set. Let $f_1:\mathbb{A}\rightarrow \mathbb{D}$ be a function such that, for every finite set $F\in\mathbb{A}$, $f_1(F)\geq x$ for every $x\in F \cap \mathbb{D}$. This can be done using definable choice because $\mathbb{D}$ is directed.

Try to define $f_2$ analogously and prove that this works.

$\endgroup$
  • $\begingroup$ Excellent! BTW, I think it would be better to define $\mathbb A:=\left\{X\cup Y|\ X\ \text{is a finite subset of } \mathbb D,\ Y\ \text{is a finite subset of } \mathbb E\right\}, $ in order to avoid the case $F\cap \mathbb D=\emptyset$ or $F\cap\mathbb E=\emptyset. $ $\endgroup$ – painday Dec 25 '18 at 4:37
  • 1
    $\begingroup$ For my supervisor zero is still a finite cardinality. Anyhow the definition I gave still works if $F \cap \mathbb{D}$ is empty. In that case $f_1(F)$ would just be any element in $\mathbb{D}$. $\endgroup$ – Anguepa Dec 26 '18 at 17:55

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.