0

I'm trying to prove the identity $$(\cup_iA_i)\times(\cup_jB_j)=\cup_{i,j}(A_i\times B_j)$$ if $\{A_i\}, \{B_j\}$ are families indexed by $I, J$ respectively. Now the logic is rather straightforward, but I'm not sure how to amend for the possibility of $I=\emptyset$ or $J=\emptyset$. I tried to let $\cup_iA_i=A$ if $I=\emptyset$ and similar improviso is made for $\cup_jB_j$ too but that doesn't seem to sustain the identity when only one of $I$ or $J$ is empty. Anyone could give me a nudge forward? Thanks!

Macrophage
  • 1,597

1 Answers1

1

Forgive me if i misunderstand, but surely the question supposes that I and J are non-empty families. If the question actually allows for an empty indexing set, then you could argue as follows: $x \in \bigcup_{\gamma \in I} A_\gamma$ would imply that there exists a $\gamma$ in $I$ such that $x \in A_\gamma$, but since $I$ is empty such a $\gamma$ does not exist, and so that would be an empty set by definition. You could then take it from there to do some sort of double inclusion, explaining the special case of taking the cartesian product of a set with an empty set If you are interested in a little more detail of how an empty indexing set might work, i will link this question

Marat Aliev
  • 131