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.
- 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
- 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