# Lesson 17 Solutions

Exercise 1) 1) Let $A=\{1, 2\}, B=\{a, b\}$, and $C=\{3,4\}$.  Then the Cartesian product $A\times B$ of A and B is a perfectly good set, and so we can take its Cartesian product with C, denoted by $(A\times B)\times C$.  How many elements are in $(A\times B)\times C$?.

Solution:  The quick answer is that there are 8 elements in $(A\times B)\times C$.  The longer answer would involve an explanation as to why this is so, so let me provide that now.  We already know that the Cartesian product of a set with N elements and one with M elements is a set with $N\times M$ elements.  Accordingly, we can simply view $(A\times B)\times C$ as the Cartesian product of a 4-element set with a 2-element set.  This is because $(A\times B)\times C$ is the Cartesian product of $A\times B$, which has 4 elements (since it is itself the Cartesian product of two 2-element sets), with C, which has 2 elements.

To be even more explicit, and to check that we’re right, we can simply write out all of the elements in $(A\times B)\times C$.  To do this, we first need all of the elements in $A\times B$.  These are simply $(1, a), (1, b), (2,a), \mathrm{\ and\ }(2,b)$.  Thus, simply inserting these 4 elements into the “first slot” of the pairs in the set $(A\times B)\times C$, we have the 8 elements $((1,a), 3), ((1, a), 4), ((1,b),3), ((1,b),4), ((2,a),3), ((2,a),4), ((2,b),3), \mathrm{\ and\ }((2,b),4)$.

Exercise 2)  Let $A=\{1, 2, 3\}$.  How many elements are in $A\times A$?

Solution: After doing the above exercise, I think this one is rather clear, the main point of it being to point out that we can indeed take Cartesian products of sets with themselves (just as we can with unions and intersections).  Thus, since A is a 3-element set, the Cartesian product of A with itself is a 9-element set.  That’s the answer.  To be more clear, however, we can list these 9 elements: $(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2),\mathrm{\ and\ } (3, 3)$.  Note, however, that while we can take unions, intersections, and Cartesian products of sets with themselves, the results of doing so are vastly different.  The union and intersection of any set with itself are both simply the original set (as you can check), whereas the Cartesian product of a set with itself is vastly different from the original set.  The difference lies in the fact that the elements in the Cartesian product are pairs of elements in the original set, and not the elements themselves.  Note also that $(1, 2)$ and $(2, 1)$ are different elements, since the pairs involve picking the “1” and “2” from different copies of A.  We can think of this as picking the elements in a different order: one involves picking 1 first and 2 second, and the other involves picking 2 first and 1 second, thus giving rise to different pairs.

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