On 11 Aug 2009, at 15:32, Mirek Dobsicek wrote:
>
>
>> 3) compute { } ^ { } and card({ } ^ { })
>
>> If card(A) = n, and card(B) = m. What is
>> card(A^B)?
>
> I find it neat to write | {} ^ {} | = | { {} } | = 01 :-)
You will make panic those who are not familiar with symbols!
>
> It's almost like ASCII art. Just wanted to signal that I'm following.
Thank for telling me.
OK, people, good time to solve the problems. Please don't read this
post, unless you find it is the good time for you to do some math. If
not, postpone until a good time. Technical posts have to be studied,
not read. Take the time needed. Tell me if I am too quick.
The solution of "3)" has been given.
Let us look at:
>> If card(A) = n, and card(B) = m. What is
>> card(A^B)?
card(A) = n
This means A is a finite set with n elements.
card(B) = m
This means B is a finite set with m elements.
Let us simplify by supposing that m = 3, and n = 2. Hoping that the
reasoning done for finding the solution on the particular case will
inspire the reasoning for finding the solution in the general case.
Let us imagine that A is the set {a, b, c}, with its three elements,
and that B is the set {1, 2}, with its two elements.
And let us try now to remember what is the question.
The question is: what is card(A^B)?
Well, card(A^B) is the number of elements of A^B. By definition of
the cardinal.
What is A^B?
Well, A^B is the set of functions from B to A. By definition of set
exponentiation.
Well, if the question was just "what is card(A^B)?", this would
provide the best solution, or the best note if you want. But the
teacher provided the information that A has n elements, and that B has
m elements, and intuitively we can bet that the number of functions
from a set to another can depend on the number of elements of each
sets involved, so that "what is card( A^B)?" meant probably how to
compute card(A^B) in function of card(A) and card(B).
Ok, we decided to look on the particular case with A = {a, b, c}, and
B = {1, 2}.
A^B = the set of all functions from B to A.
That is the set of functions from {1, 2} to {a, b, c}.
Well, let us try to find, or to build, one function from B to A.
But , here a moment of panic can occur, (empirical observation). For
the unnameable sake, what *is* a function? What is a function from B
to A.
Well, if it is an open manual home work, such panic can be eased by
looking in the math notes. You may remember the motivation or the
informal sense of what a function represents, which is a relation of
dependency, and this is in the most general sense, so that all
possible dependency are tolerated. For a function from B to A, it
means the element of A depends in function of the elements of B. Such
a dependency is well described by a couple (x, y) with x in B and y in
A.
we have (x,y) belongs-to F representing the meaning that y depends "in
the function F" of x.
Think about x as time and y as temperature.
So, a function from B to A is just a set of couples (x, y) with x in B
and y in A, with the functional restriction that x is not send to two
different values y. At each time x, you can have only one temperature y.
That is, here: a set of couples (x, y) with x in {1, 2} and y in {a,
b, c}, and such that if (1, x) belongs-to F, no other (1, y) belongs
to F.
Let us build one function from {1, 2} to {a, b, c}.
OK, 1, from B, can determine what in A ? Well, we have three
possibilities a, b and c. OK, i will use my free will to decide that
for this function I want now, 1 will determine a. So I put the couple
(1, a) in the function.
At this stage, the "function" looks like {(1, a)}.
Finished?
No, a function from a set to another one gives a values, outcomes,
outputs for all elements of its domain. I have to say what is
determine by 2, in B. OK, I will use my free will again, and decide to
add the couples (2, a).
At this stage, the function looks like {(1, a) (2, a)}.
Finished?
Yes.
We do have a function from B to A. The set {(1, a) (2, a)} describes
completely a function from B to A, a so-called "constant function".
think of 1 and 02 as moment of times, and think of a, b, c, as possible
temperature. The function {(1, a) (2, a)} describe a case here the
temperature is constant and equal to a.
Finished? No, we have to find all functions from B to A. All functions
from {1, 2} to {a, b, c}.
Well actually, we need to find only the number of such functions. For
1 I have three choices, then for 2, I have still three choices, and
the choices are independent, so that for each choice the remaining
three choice will lead to distinct functions, this make 03 X 3
functions = 09 functions:
{(1, a) (2, a)}
{(1, a) (2, b)}
{(1, a) (2, c)}
{(1, b) (2, a)}
{(1, b) (2, b)}
{(1, b) (2, c)}
{(1, c) (2, a)}
{(1, c) (2, b)}
{(1, c) (2, c)}
so A^B = {{(1, a) (2, a)}, {(1, a) (2, b)}, {(1, a) (2, c)}, ... ,
{(1, c) (2, c)}}, and card(A^B) = 9. In this case. This give all the
way the a, b, c can depend on 1 and 2.
I stop here. I let you train on the following question:
How many functions from {a, b, c} to {1, 2}?
How many functions from {1, 2, 3, 4, 5} to {a, b, c, d, e}?
What is the general solution, in term of cardinal n and m of the sets
involved ? (the original question).
Take your time, and ask any question. This is the type of stuff rather
easy for exact scientists, and rather new for those who buried math in
their unconscious in high school, so take each your own time.
I hope I am not too long. We will see many many examples of functions.
Bruno
http://iridia.ulb.ac.be/~marchal/
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Received on Tue Aug 11 2009 - 21:36:05 PDT