IV. Product of Sets: Exercises

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


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.


Nalin Pithwa

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