# Interactive Interlude 2: My Ideas

Idea 1) The union forms the set of elements that are in either original set, and the intersection forms the set of elements that are in both sets.  To me, the next natural construction would be to consider the set of elements that are in exactly one of the original sets.  I.e., the set of elements that are in one and only one of the original sets—if it’s in both, it doesn’t count.  Let me call this construction XOR (computer scientists will know why) which is read off as “exclusive or”.  Then, I’d write A XOR B for the new set.  For example, if $A=\{1, 2, 3, 4\}$ and $B=\{1, 4, 5, 6\}$, then $A\ XOR\ B=\{2, 3, 5, 6\}$.  This is because 1 and 4 show up in both sets, which is no bueno for my construction, and all other elements in the universe show up nowhere in A or B and so accordingly don’t show up in A XOR B.

Idea 2) We introduced the Cartesian product as a way to describe the set of “pairs” of elements, where each element in the pair comes from one of the two sets that we’re Cartesian product-ing.  Let me now define a bunch of other similar constructions, involving “triples” or “quadruples” of elements.  For example, if A and B are sets, I can define the set of all elements of the form $(a, a', b)$ where $a,a'\in A$ and $b\in B$.  In other words, we’re just making a “triple” such that a “pair” comes from A itself, and the last slot comes from B.  As it stands, however, this construction would be the same as just taking the Cartesian product of A with itself, and then taking the product of this new set with B (can you see why?).  In order to make this a genuinely new construction, we can put new requirements on this set.  For example, we could make it so that we only include the elements $(a, a', b)$ where the first slot does not equal the second slot (this make sense, since the first and second slot both come from the same set and therefore we know how to ask if they’re the same or not).

We could expand on this by considering quadruples and even higher “order” constructions, like the set of elements $(a, a', a'', b, b', b'')$ such that $a, a', a''\in A$ and $b, b', b''\in B$ where we require that no two elements coming from the same set are equal.  Some interesting things happen here, though.  For example, in this case (where the elements have 6 slots), if A and B both have less than 3 elements, then this type of product is in fact the empty set!  This is because we can’t find three elements that are all not equal to each other to fill the three slots for allocated for each set, simply because the sets don’t have enough distinct elements in them.

Consider the set with four slots, where each component set is allocated two slots, and where the elements need to be distinct in the slots (the set of elements $(a, a', b, b')$ with $a, a'\in A$ and $b, b'\in B$ and where the first slot can’t equal the second slot, and the third slot can’t equal the fourth slot).  This  product of two 2-element sets has exactly four elements.  To see this, let $A=\{1, 2\}$ and $B=\{3, 4\}$.  Then the 4-element product is $\{(1, 2, 3, 4), (1, 2, 4, 3), (2, 1, 3, 4), (2, 1, 4, 3)\}$.

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On to Lesson 19

### 2 Responses to Interactive Interlude 2: My Ideas

1. Joe says:

Is your ‘idea 1’ equivalent to the notion of “set symmetric difference”? Thanks for these lessons, am enjoying them.

• TrueBeautyOfMath says:

Yep! It is, good call. Sorry, I should have mentioned that in there—I’ll add it in when I can. And I’m glad you’re enjoying the lessons! Please don’t hesitate to keep asking questions.