Cartesian products involving empty sets

Question

Let $(\Omega_i)_{i\in I}$ be a nonempty family of sets. For each $i \in I$, let $\mathcal{E}_i$ be a set of subsets of $\Omega_i$. Let

$$ \coprod_{i\in I} \mathcal{E}_i:=\bigg\{\prod_{i\in I}A_i : A_i\in\mathcal{E}_i \cup\{\Omega_i\} , A_i\neq \Omega_i \text{ for finitely many }i\in I \bigg\}$$

Let $(I)_{\lambda \in \Lambda}$ be a partition of $I$ into disjoint nonempty sets. I am trying to prove that

$$\coprod_{i\in I} \mathcal{E}_i= \coprod_{\lambda \in \Lambda} \bigg( \coprod_{i\in I_\lambda} \mathcal{E}_i\bigg)$$

when identifying sets of the form $\prod_{i\in I} A_i$ $(A_i\subset\Omega_i)$ with $\prod_{\lambda \in \Lambda} \big( \prod_{i\in I_\lambda} A_i\big)$. I already showed $\coprod_{i\in I} \mathcal{E}_i \subset \coprod_{\lambda \in \Lambda} \bigg( \coprod_{i\in I_\lambda} \mathcal{E}_i\bigg)$, but I am not sure the reverse inclusion holds if $\Omega_i =\emptyset$ for some $i$.

Is this reasoning correct? Thanks a lot for your help.

Answer 1

$\Omega_i = \emptyset$ does not cause a problem, both sides will just be empty.

Your definition of $\coprod_{\lambda \in \Lambda} \bigg( \coprod_{i\in I_\lambda} \mathcal{E}_i\bigg)$ is incomplete, as you haven't defined an $\Omega_{I_\lambda}$ for each member of $\coprod_{i\in I_\lambda} \mathcal{E}_i$ to be a subset of. If we put that to be $\bigcup \bigg( \coprod_{i\in I_\lambda} \mathcal{E}_i\bigg)$ then your equality does hold.

To show the reverse inequality note that any member $\prod_{\lambda \in \Lambda} B_\lambda$ of $\coprod_{\lambda \in \Lambda} \bigg( \coprod_{i\in I_\lambda} \mathcal{E}_i\bigg)$ will have, for all but finitely many $\lambda$, $B_\lambda = \Omega_{I_\lambda} = \prod_{i\in I_\lambda} \Omega_i$. So each member is of the form $\prod_{\lambda \in \Lambda} \big( \prod_{i\in I_\lambda} A_i\big)$ with $A_i \neq \Omega_i$ for only finitely many $i$ from only finitely many $I_\lambda$, and so it identifies with $\prod_{i\in I} A_i$ for only finitely many $A_i \neq \Omega_i$ overall.

Answer 2

Your assumption, condition, prerequisite that:

Let $\Omega_{i_{i \epsilon I}}$ be a nonempty family of sets.

implies, garanties that

Ω=∅ for some .

is not possible. There exists no such Ω=∅ for all ∈ that Ω=∅!

$$\not \exists \Omega_{i_{i \epsilon I}} : \Omega_{i}=∅$$

or

$$\forall \Omega_{i_{i \epsilon I}} : \Omega_{i}\not = ∅$$

Use this argument and You are done. The condition is more than that I in nonempty. This website provides some inside into the meaings of the condition: MathWorld Family. The article contains a passage that explains the details: