# Prob. 3, Sec. 4, in G.F. Simmons' INTRO TO TOPOLOGY & MODERN ANALYSIS: A ring of subsets of $X \times Y$

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?

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.
• 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? – Saaqib Mahmood May 20 '19 at 11:35
• It is true in general that $$\left[\bigcup_{i=1}^m A_i\right]-B=\bigcup_{i=1}^m(A_i-B)$$ and so $$\left[ \bigcup_{i=1}^m (A_i \times B_i) \right] - \left[ \bigcup_{j=1}^n (C_j \times D_j)\right]=\bigcup_{i=1}^m\left[(A_i\times B_i)-\left[\bigcup_{j=1}^n(C_j\times D_j)\right]\right]$$ – PJK May 20 '19 at 15:52
$$W:=U-(C_1\times D_1)=\bigcup_{i=1}^m\left[(A_i\times B_i)-(C_1\times D_1)\right]=$$$$\bigcup_{i=1}^m\left[(A_i-C_1)\times B_i)\cup(A_i\times(B_i-D_1)\right]=\left[\bigcup_{i=1}^m(A_i-C_1)\times B_i\right]\cup\left[\bigcup_{i=1}^mA_i\times(B_i-D_1)\right]$$ So $$W$$ belongs to the collection. Further:$$U-\bigcup_{j=1}^n \left( C_j \times D_j \right)=W-\bigcup_{j=2}^n \left( C_j \times D_j \right)$$
• 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. – drhab May 20 '19 at 12:23
• 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. – Saaqib Mahmood May 20 '19 at 13:24