# IV. Product of Sets: Exercises

Reference: Topology and Modern Analysis, G F Simmons, Tata McGraw Hill Publications, India.

Problems:

I) The graph of a mapping $f: X \rightarrow Y$ is a subset of the product $X \times Y$. What properties characterize the graphs of mappings among all subsets of $X \times Y$?

Solution I: composition.

II) Let X and Y be non-empty sets. If $A_{1}$ and $A_{2}$ are subsets of X, and $Y_{1}$ and $Y_{2}$ are subsets of Y, then prove the following

(a) $(A_{1} \times B_{1}) \bigcap (A_{2} \times B_{2}) = (A_{1} \bigcap A_{2}) \times (B_{1} \bigcap B_{2})$

(b) $(A_{1} \times B_{1}) - (A_{2} \times B_{2}) = (A_{1}-A_{2}) \times (B_{1}-B_{2}) \bigcup (A_{1} \bigcap A_{2}) \times (B_{1}-B_{2}) \bigcup (A_{1}-A_{2}) \times (B_{1} \bigcap B_{2})$

Solution IIa:

TPT: $(A_{1} \times B_{1}) \times (A_{2} \times B_{2}) = (A_{1} \bigcap A_{2}) \times (B_{1} \bigcap B_{2})$

This is by definition of product and the fact that the co-ordinates are ordered and the fact that $A_{1} \subseteq X$, $A_{2} \subseteq X$, $B_{1} \subseteq Y$, and $B_{2} \subseteq Y$.

Solution IIb:

Let $(a_{i}, b_{j}) \in A_{1} \times B_{1}$, but $(a_{i}, b_{j}) \notin A_{2} \times B_{2}$. So, the element may belong to $(A_{1}-A_{2}) \times (B_{1} - B_{2})$ or it could happen that it belongs to $A_{1} \times A_{1}$, but to $(B_{1}-B_{2})$ (here we need to remember that the element is ordered); so, also it could be the other way: it could belong to $(A_{1}-A_{2})$ but to $(B_{1} \bigcap B_{2})$ also. The same arguments applied in reverse establish the other subset inequality. Hence, done.

III) Let X and Y be non-empty sets, and let A and 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 $A \in \bf{A}$ and $B \in \bf{B}$ is a ring of subsets of $X \times Y$.

Solution III:

$\bigcup_{i=1}^{n}A_{i} \times B_{i} = \bigcup_{i=1}^{n}A_{i} \times \bigcup_{i=1}^{n}B_{i}$.

From question IIb above, the difference of any two pairs of sets is also in $X \times Y$.

Hence, done.

Regards,

Nalin Pithwa

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