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Here is Prob. 3, Sec. 4, in the book Introduction to Topology & Modern Analysis by George F. Simmons:

Let $X$ and $Y$ be non-empty sets, and let $\mathscr{A}$ and $\mathscr{B}$ be rings of subsets of $X$ and $Y$, respectively. Show that the class of all finite unions of sets of the form $A \times B$ with $A \in \mathscr{A}$ and $B \in \mathscr{B}$ is a ring of subsets of $X \times Y$.

And, here is Prob. 4, Sec. 2, in that very book:

A ring of sets is a non-empty class $\mathscr{A}$ of sets such that if $A$ and $B$ are in $\mathscr{A}$, then $A \Delta B$ and $A \cap B$ are also in $\mathscr{A}$. Show that [a ring of sets} $\mathscr{A}$ must also contain the empty set, $A \cup B$, and $A - B$. Show that if a non-empty class of sets contains the union and difference of any pair of its sets, then it is a ring of sets. . . .

I also know that, for any subsets $A_1$ and $A_2$ of $X$ and for any subsets $B_1$ and $B_2$ of $Y$, the following hold: $$ \left( A_1 \times B_1 \right) \cap \left( A_2 \times B_2 \right) = \left( A_1 \cap A_2 \right) \times \left( B_1 \cap B_2 \right). \tag{1} $$ $$ \left( A_1 \times B_1 \right) - \left( A_2 \times B_2 \right) = \big[ \left( A_1 - A_2 \right) \times \left( B_1 - B_2 \right) \big] \cup \big[ \left( A_1 \cap A_2 \right) \times \left( B_1 - B_2 \right) \big] \cup \big[ \left( A_1 - A_2 \right) \times \left( B_1 \cap B_2 \right) \big]. \tag{2} $$ $$ \left( A_1 \cup A_2 \right) \times \left( B_1 \cup B_2 \right) = \big[ A_1 \times B_1 \big] \cup \big[ A_1 \times B_2 \big] \cup \big[ A_2 \times B_1 \big] \cup \big[ A_2 \times B_2 \big]. \tag{3} $$

My Attempt:

Suppose that $U$ and $V$ be any two sets of our collection. Then there exist sets $A_1, \ldots, A_m, C_1, \ldots, C_n$ in $\mathscr{A}$ and sets $B_1, \ldots, B_m, D_1, \ldots, D_n$ in $\mathscr{B}$ such that $$ U = \bigcup_{i=1}^m \left( A_i \times B_i \right), \qquad \mbox{ and } \qquad V = \bigcup_{j=1}^n \left( C_j \times D_j \right). $$

We need to show that both $U \cup V$ and $U - V$ are also in our collection. Or, we need to show that both $U \Delta V$ and $U \cap V$ are also in our collection.

What next? How to proceed from here?

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It is clear that $U\cup V$ is in the collection. To prove that $U-V$ is also in the collection, first note that for any sets $A, B$ and $C$ $$A-(B\cup C)=(A-B)\cap (A-C).$$ So $$(A\times B)-[(C_1\times D_1)\cup(C_2\times D_2)]=[(A\times B)-(C_1\times D_1)]\cap[(A\times B)-(C_2\times D_2)].$$ By (1) and (2), this set also belongs to the collection. Now an application of induction will prove the result.

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  • $\begingroup$ what you've shown leads to the following. For any $n \in \mathbb{N}$, the set $$(A\times B) - \left[ \bigcup_{j=1}^n (C_j \times D_j)\right]$$ also belongs to our collection. Am I right? How then to use this to show that the set $$\left[ \bigcup_{i=1}^m (A_i \times B_i) \right] - \left[ \bigcup_{j=1}^n (C_j \times D_j)\right]$$ also belongs to our collection? Can you please expand upon this as well? $\endgroup$ – Saaqib Mahmood May 20 at 11:35
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    $\begingroup$ It is true in general that $$\left[\bigcup_{i=1}^m A_i\right]-B=\bigcup_{i=1}^m(A_i-B)$$ and so $$\left[ \bigcup_{i=1}^m (A_i \times B_i) \right] - \left[ \bigcup_{j=1}^n (C_j \times D_j)\right]=\bigcup_{i=1}^m\left[(A_i\times B_i)-\left[\bigcup_{j=1}^n(C_j\times D_j)\right]\right]$$ $\endgroup$ – PJK May 20 at 15:52
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$$W:=U-(C_1\times D_1)=\bigcup_{i=1}^m\left[(A_i\times B_i)-(C_1\times D_1)\right]=$$$$\bigcup_{i=1}^m\left[(A_i-C_1)\times B_i)\cup(A_i\times(B_i-D_1)\right]=\left[\bigcup_{i=1}^m(A_i-C_1)\times B_i\right]\cup\left[\bigcup_{i=1}^mA_i\times(B_i-D_1)\right]$$ So $W$ belongs to the collection. Further:$$U-\bigcup_{j=1}^n \left( C_j \times D_j \right)=W-\bigcup_{j=2}^n \left( C_j \times D_j \right)$$

This can be repeated.

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  • $\begingroup$ I've just edited your answer. Can you please check and comment on whether you agree or not? And, I'm unable to make sense of the very last equality in your answer. So can you please elaborate? $\endgroup$ – Saaqib Mahmood May 20 at 11:29
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    $\begingroup$ We have $(A_i-C_1)\times(B_i-D_1)\subseteq(A_i-C_1)\times B_i$ so there is no need to mention $(A_i-C_1)\times(B_i-D_1)$ separately. In my original answer I left that set out. It is proved that $W$ belongs to the collection (do you agree?) From $W=U-(C_1\times D_1)$ it follows directly that $W-\bigcup_{j=2}(C_j\times D_j)=U-C_1\times D_1-\bigcup_{j=2}(C_j\times D_j)=U-\bigcup_{j=1}(C_j\times D_j)$. The union $\bigcup_{j=2}(C_j\times D_j)$ has one set less in the union. This makes evident that we can prove by induction on $n$. I intend to roll back your edit, but will first wait for a reaction. $\endgroup$ – drhab May 20 at 12:23
  • $\begingroup$ yes, I agree that (1) my edit is superfluous although I just used what the preceding problem in the book has asked for the proof of and (2) $W$ does belong to our collection. So please feel free to roll back, as you please. $\endgroup$ – Saaqib Mahmood May 20 at 13:24

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