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Suppose $I \neq \emptyset $. Prove that for any indexed family of sets {${A_i|i\in I}$} and any set $B$, $(\cap_{i\in I}A_i)$ x $B$ = $\cap_{i\in I}(A_i$ x $B)$. Where in the proof does the assuption that $I \neq \emptyset $ get used?

$\rightarrow$

$(\cap_{i\in I}A_i)$ x $B$ =

$x \in \cap_{i\in I}Ai$ $\land y\in B$ =

$\forall_{i\in I}(x\in A_i) \land y \in B$ =

$\forall_{i\in I}(x\in A_i) \land \forall_{i\in I}(y\in B)= $

$\cap_{i\in I}(A_i$x$B)$

$\leftarrow$

similar.

Where in the proof should I use $I \neq \emptyset $?

Josue
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    The "empty intersection" (i.e. $\cap_{i\in I}$ where $I=\varnothing$) does not make sense here, since the $A_i$ do not all necessarily live inside some larger set $S$. See: https://math.stackexchange.com/questions/370188/empty-intersection-and-empty-union Edit: really what you get is the "universal set" (eek!) which doesn't exist in ZFC (google "Russell's Paradox") – morrowmh Jun 06 '21 at 03:30
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    I believe you can drop the assumption that $I\not=\varnothing$ as long as you add the assumption that there exists a set $S$ such that $A_i\subseteq S$ for all $i\in I$. Then if $I=\varnothing$ you have $\cap_{i\in I}A_i=S$, and similarly $\cap_{i\in I}(A_i\times B)=S\times B$. – morrowmh Jun 06 '21 at 03:37
  • @MichaelMorrow thank you. – Josue Jun 06 '21 at 03:40

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