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I am trying to show that if $\epsilon_1$ and $\epsilon_2$ are 2 elementary families of sets in $X_1$ and $X_2$ respectively then $$\epsilon_1 \times \epsilon_2=\{E_1 \times E_2 | E_1 \in \epsilon_1, E_2 \in \epsilon_2\}$$ is an elementary family of sets on $X_1 \times X_2$.

My book, Folland, defines an elementary family to be a collection $\epsilon$ of subsets of $X$ such that:

1) The empty set is in $\epsilon$

2) If $E,F \in \epsilon$, then $E \cap F \in \epsilon$

3) If $E \in \epsilon$, then $E^c$ is a finite disjoint union of members of $\epsilon$

For this problem I am trying to show the following:

1) The empty set is in $\epsilon_1 \times \epsilon_2$ is clear since it is in both $\epsilon_1$ and $\epsilon_2$.

I am not sure about 2) and 3)...

2) I want to show that if $E,F \in \epsilon_1 \times \epsilon_2$, then $E \cap F \in \epsilon_1 \times \epsilon_2$

We know that there exists $E_i,E_m \in \epsilon_1$ and $E_j,E_n \in \epsilon_2$ so that $E=E_i \times E_j$ and $F=E_m \times E_n$. Then, $E \cap F=(E_i \times E_j) \cap (E_m \times E_n)$ where both terms in parentheses are in $\epsilon_1 \times \epsilon_2$ by definition. So we have that $E \cap F \in \epsilon_1 \times \epsilon_2$. I am wondering if this is correct...

3) I want to show that If $E \in \epsilon_1 \times \epsilon_2$, then $E^c$ is a finite disjoint union of members of $\epsilon_1 \times \epsilon_2$.

Again, we know that there exists $E_i \in \epsilon_1$ and $E_j \in \epsilon_2$ so that $E=E_i \times E_j$. I then said that $E^c=(E_i \times E_j)^c = (E_i^c \times X_1) \cup (E_j^c \times X_2)$ but I am stuck because $(E_i^c \times X_1)$ and $(E_j^c \times X_2)$ are not disjoint as far as I understand...

Thank you for any help you can give me.

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    $\begingroup$ In 2, the point is that $E\cap F=(E_i\cap E_m)\times (E_j\cap E_n)$. $\endgroup$ – Angina Seng Aug 27 '18 at 18:19
  • $\begingroup$ @LordSharktheUnknown why is that? Shouldn't E be made up of some $E_i \in X_1$ and some $E_j \in X_2$? $\endgroup$ – MathIsHard Aug 27 '18 at 18:24
  • $\begingroup$ Oh I think I misread what you put. Thank you I see what you mean now. :) $\endgroup$ – MathIsHard Aug 27 '18 at 18:26
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For 3, let $E\in\varepsilon_1$ and $E'\in\varepsilon_2$. Then $X_1=E_0\cup E_1\cup\cdots\cup E_m$ where this is a disjoint union of elements of $\varepsilon_1$ and $E_0=E$. Likewise $X_2=E_0'\cup E_1'\cup\cdots\cup E_n'$ where this is a disjoint union of elements of $\varepsilon_2$ and $E_0'=E'$.

Then $(E\times E')^c$ is the disjoint union of the $E_i\times E_j'$ for $(i,j)\ne(0,0)$.

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  • $\begingroup$ Oh thank you very much. I appreciate the help. :) I see now that my approach wasn't quite correct. $\endgroup$ – MathIsHard Aug 27 '18 at 18:28

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