**Exercise 1) Let . How many elements are in ? How many elements are in ? Are the sets and equal? (This is a subtle question, so if you’re a bit stumped have no worries, I’ll try to explain fully in the solutions).**

Solution: Recall from the lesson that the disjoint union of a set with N elements and a set with M elements has N+M elements. Thus, the disjoint union of A and B has 4 elements (since they both have 2 elements), and so the disjoint union of “A disjoint union B” and C (where “A disjoint union B” is a single set!) has 4+2=6 elements. I.e., has 6 elements. The exact same logic applies to show that also has 6 elements.

Now in order to answer the last question in this exercise, we need to get a bit more detailed. Let’s, for the sake of argument, write out all of the 6 elements in . I am going to do this by strictly sticking to the notation developed in this lesson, and I’ll warn you now that this will get ugly. But remember, it’s **just notation**, and we should never let notation scare us! All we need to do is take a deep breath and remember what the notation means.

To do this, we first need to write out the elements in . From what we learned in the lesson, we see that we have , where we put the element in the “first slot” and the set that “labels” the element in the second slot. Now, to write out the elements in , we simply do this again. Namely, we write out the elements of each of these two sets in the first slot, with the title of the set “labeling” it in the second slot. Thus we have

If you look closely, you’ll see that we did nothing but take all of the elements in one set and put them in the first slot (only now for the set the elements themselves already have one set of parentheses (hence the nested parentheses)) while labeling them with the name of the set in the second slot. The elements in C don’t already have parentheses (and therefore they don’t have nested parentheses) because we have not already disjoint union-ed them with anything.

Viewing things in this way, it becomes clear that the answer to the last question is no, simply because the elements that we create when making the set are different than those that we create when we make the set . In particular, we have that

and these elements are clearly different than those in !

Remember, all of these crazy looking sets are just the inevitable results of the basic notation that we set up in the lesson, and are therefore nothing to be scared of.

**Exercise 2) Let A and B be as above in Exercise 1). How many elements are in the set ? (Recall that this means “the disjoint union of the set ‘A union B’ with the set C”).**

Solution: Compared to the first exercise, this one’s a bit easier. Again, the main fact is that the disjoint union of a set with N elements and a set with M elements has N+M elements, so all we need to do is see how many elements the “component” sets have. Well C clearly has two elements, so all we need to do is ask about . Remember, though, that the **regular** union **does not** distinguish between repeated elements, so only has 3 elements because “2” is in both sets. Thus we’re taking the disjoint union of a set with 3 elements and a set with 2 elements, so the result will have 5 elements. Done!

On to the Interlude

On to Lesson 19