Re: The seven step series

```On 28 Jul 2009, at 17:36, m.a. wrote:

> Bruno,
>              I have searched my notes for an exposition of BIJECTION
> and found only one mention in an early email which promises to
> define it in a later lesson. Do you have a reference to that lesson
> or perhaps an instant explanation of it? Thanks,
>
> Chief
>  Ignoramus```
```

I hope you are not stuck by that, given that the cartesian product
does not rely on the understanding of "bijection".

An instant explanation of bijection is this:

Suppose you have two sets A and B, and you would like to know if they
have the same number of elements. For example:

A = {a,b,c,d,e,f,g}

and

B = {1,  2,  3,  4,  5,  6,  7}

Suppose that you cannot count. You forget the lessons for counting!

But you have ficelles, I mean ropes, cords, or strings.

So you can line up the two sets, and try to attach to each elements of
A a piece of rope, and joint them to one element of the set B.
IF you succeed doing that, and respecting the one-one or 1-1link, and
getting all the elements of B (the "onto" condition), THEN you have
shown the existence of a bijection between the two sets.

Let us see if that work on the example.  The "----------" represent
the pieces of rope. OK?

a  ----------  1
b  ----------  2
c  ----------  3
d  ----------  4
e  ----------  5
f  ----------  6
g  ----------  7

So there is a bijection between A and B.

The bijection *is* that association, as we will defined much later(*).

Other bijections can exist between A and B, like

a  ----------  7
b  ----------  2
c  ----------  3
d  ----------  4
e  ----------  5
f  ----------  6
g  ----------  1

It is enough that one exist, to conclude the sets have the same
"number of elements", or same cardinal.

Convince you that if two sets have different number of elements, there
is no bijection in between. It has to be 1-1, and "onto" (no missing
element).

Exercise (but no hurry):
Verify if you can see some bijections existing between the powerset of
a set with 2 elements, and B_2. The same for the powerset of a set
with 3 elements, and B_3.

Bruno

!*) Oh! I can give you the particular mathematical bijection, existing
between A and B, given that I have already define the notion of couple.
It is the set of couples representing naturally those rope association:

{(a,1), (b,2) (c,3) (d,4), (e,5), (f,6), (g,7)}.    Take it easy, and
be sure you have read what I say about the couples before.

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

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