Algebra Know-It-ALL

infinitely many elements. Amazingly enough, you can pair the elements of both sets

off one-to-one! The mechanics of this goes a little beyond the scope of this chapter,

but you can get an idea of how it works if you divide every element of Weven by 2, one

at a time, and then write down the first few elements of the resulting set. When you

do that, you get

{0/2, 2/2, 4/2, 6/2, 8/2, 10/2, ...}

But that’s exactly the same as W, because when you perform the divisions, you get

{0, 1, 2, 3, 4, 5, ...}

This is one of the strange things “infinity” can do. You can take away every other element

of a set that has infinitely many elements and that can be written out as an “implied list,”

and the resulting set is exactly the same “size” as the original set.

7. Let’s write out the sets again here as “implied lists”:

A= {1, 1/2, 1/3, 1/4, 1/5, 1/6, ...}

G= {1, 1/2, 1/4, 1/8, 1/16, 1/32, ...}

It’s not hard to see that set G contains all those elements, but only those elements, that

belong to both sets. Therefore

A∩G=G

If you start with set A and then toss in all the elements of G, you get the same set A (with

certain elements listed twice, but they can count only once). That means set A contains

precisely those elements that belong to one set or the other, or both, so

A∪G=A

8. First, isolate all the individual elements of the set {1, 2, 3}. They are 1, 2, and 3. Then
   start the list of subsets by putting down the null set, which is a subset of any set. Then
   assemble all the sets you possibly can, using one or more of the elements 1, 2, 3, and list
   them. You should get

∅

{1}

{2}

{3}

{1, 2}

{1, 3}

{2, 3}

{1, 2, 3}

Chapter 2 591