2
$\begingroup$

Following is similar to my earlier questions, but try to understand them from category theory. Mariano said it was possible in a comment, but I don't know how.

An object $X$ is the product of a family $\{X\}_i, i \in I$ of objects iff there exist morphisms $\pi_i : X \to X_i$, such that for every object $Y$ and a $I$-indexed family of morphisms $f_i : Y \to X_i$ there exists a unique morphism $f : Y \to X$ such that the following diagrams commute for all $i \in I$:

enter image description here

When applying that definition to set systems $(E_i, \mathcal{B}_i), i \in I$ of the same type $\theta$, to get their product set system $\mathcal{B}$ on $E= \prod_{i \in I} E_i$, I was wondering how that leads to $$\mathcal{B}=\theta \left(\left\{\text{$\prod_{i \in I}B_i$, where $B_i \in \mathcal{B}_i, B_i=E_i$ for all but finitely many $i \in I$}\right\}\right),$$ where $\theta(\cdot)$ means taking the smallest set system of the type $\theta$ containing $\cdot$.

Can it be $$\mathcal{B}=\theta\left(\left\{\text{$\prod_{i \in I}B_i$, where $B_i \in \mathcal{B}_i, B_i=E_i$ at least for all but one $i \in I$}\right\}\right),$$ instead?

Examples of product systems are product sigma algebras, product topology, and product of set systems defined to be closed under union only.

Thanks and regards!

$\endgroup$
1
$\begingroup$

Yes. It should originally be rather the "all but one $i$" version as you suggested, because the point is that each individual projection $\pi_i$ has to be a morphism in the given category.

Therefore, in case of systems closed under union only, I have doubts that the former means the same, because it is an application of finite intersection only.

$\endgroup$

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.