# Re: The seven step series

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Hi,```
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OK, I will come back on the square root of 2 later.

We have talked on sets.

Sets have elements, and elements of a set define completely the set,
and a set is completely defined by its elements.

Example: here is a set of numbers {1, 2, 3}
and a set of sets of numbers {{1, 2}, {3}, { }}.

We can do some operations, like their union, or their intersection.
Examples:
{1,2,3} union {3,4,5} = {1,2,3,4,5},
{1,2,3} intersection {3,4,5} = { }.

We can verify if some relation hold for them, like equality, or
inclusion.
{1, 2} = {2, 1, 1}   (yes!)
{1, 2} included-in {3, 2, 1}
{1, 2} not-included-in {1, 3}

We can compute their powerset.
Powerset {1, 2} = {{ }, {1}, {2}, {1, 2}}

We have discovered SBIJECTION between powersets of a set with cardinal
n, and the set of binary strings of length n.
And we have presented reasons for the existence of a bijection between
the powerset of N = {0, 1, 2, ...} and the set of infinite binary
strings.

OK?

Today, I suggest we look at two new operations on sets. The product of
sets, and the exponentiation of sets. Well, I will probably do only
the product today.

First I have to introduce a new, well actually *very* well known,  and
absolutely important, notion: the couple.

A couple is when there is two things, but with some order. It looks
like a pair, but the order counts.

Usually a couple of things a , b is designated, in math, like this:

(a, b).

It looks like a pair {a, b}, but it is not. Indeed, {a, b} = {b, a},
but the couple (a, b) is NOT equal to the couple (b, a).

When are two couples (a, b) and (c, d) equal? Only when a = c and b = d.

Examples. the couple of number (2, 3) is not equal to the couple (3,
2), but the couple (0, 666) is equal to the couple (0, 666).

OK?

APARTE: Are couples sets? No. Nor are numbers. But yes, you can easily
represent them by sets, so we could work only with sets, but we will
not do that. Much later we will work only with numbers, in fact. The
very notion of representation will be important, though.

Now we are ready to define the so called "cartesian" product of sets.
It is indeed a cousin of Descartes' discovery that you can represent a
point of the plane by a couple of (real) numbers. I read somewhere
that Descartes discovered this by trying to describe a spider walking
on a window with squared little piece of glass. But such a
localization works also for cities like Los Angeles where you address
is something like 15th avenue 61th street. The whole field of
analytical geometry is founded on this idea.

That cartesian idea generalises on sets A and B. It is written A X B,
and it is defined by
the set of couples (x, y) such that x belongs to A, and y belongs to B.

AXB = {(x, y) such-that x belongs-to A, and y belongs to B}.
(compare with the preceding definitions).

Example:  what is the product {0, 1} X {a, b}?  Well it is the set of
all the couples made from elements of A in company of elements of  B,
and in that order, with A = {0, 1}, and B = {a, b}.
So (0, a) is in there, and there are others. The product of {0, 1)
with {a, b} is equal to

{(0,a), (0, b), (1, a), (1, b)}

The convenient usual cartesian drawing is, for AXB, with A = {0, 1},
and B = {a, b} :

a     (0, a)   (1, a)

b     (0, b)  (1, b)

0          1

A product of numbers a and b,  ab, can be conceived as the area of a
rectangle of sides a and b. Here you can see that the product of sets
AXB can fit in a rectangle when you dispose horizontally the elements
of A, and vertically the elements of B. By convention, usually A is
put horizontally, and B vertically.
But note that if the number ab is equal to the number ba, it is not
the case that the set AXB is equal to the set BXA. (0, a) does not
belong to BXA, for example.

Exercise: to the cartesian drawing for BXA.

1)
Compute
{a, b, c} X {d, e} =
{d, e} X {a, b, c} =
{a, b} X {a, b} =
{a, b} X { } =

2)
Convince yourself that the cardinal of AXB is the product of the
cardinal of A and the cardinal of B.
A and B are finite sets here. Hint: meditate on their cartesian drawing.

3) Draw a piece of NXN.

Solution and sequel tomorrow.

Any question?

Bruno

http://iridia.ulb.ac.be/~marchal/

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