# Functions / Sets / Finite Sets

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• Mar 27th 2013, 02:53 PM
emakarov
Re: Functions / Sets / Finite Sets
Well, for problem 1, you can't avoid learning about ordered pairs and Cartesian products. And it is better to read about them in a textbook written by specialists than have some amateurs explain it here.
• Mar 27th 2013, 07:19 PM
rhymin
Re: Functions / Sets / Finite Sets
An ordered pair (a, b) is a pair of mathematical objects. In the ordered pair (a, b), the object a is called the first entry, and the object b the second entry of the pair. Alternatively, the objects are called the first and second coordinates, or the left and right projections of the ordered pair. The order in which the objects appear in the pair is significant: the ordered pair (a, b) is different from the ordered pair (b, a) unless a = b.

The Cartesian product of two sets and (also called the product set, set direct product, or cross product) is defined to be the set of all points where and . It is denoted , and is called the Cartesian product since it originated in Descartes' formulation of analytic geometry. In the Cartesian view, points in the plane are specified by their vertical and horizontal coordinates, with points on a line being specified by just one coordinate. The main examples of direct products are Euclidean three-space (, where are the real numbers), and the plane ().

I dont know, but someone else explaining it might help more than that. It sounds confusing lol.
• Mar 27th 2013, 07:57 PM
Ant
Re: Functions / Sets / Finite Sets
Quote:

Originally Posted by rhymin
3) Give an example of finite sets A and B with |A|, |B| ≥ 4 and a function f: A→B such that

a) f is neither one-to-one nor onto.

Finite sets are collections of objects which are finite. |A| denotes the number of elements (objects) in the set A. So the questions asks us to find two sets, which are both have more than 4 objects in, then we have to define a function from A to B which is neither one to one nor onto.

I honestly believe that if you understand all the terms and definitions in this question then it is reasonably straightforward:

Let A = {1,2,3,4,5,6,7,8,9,10}
let B = {a,b,c,d,e}

consider the map given explicitly by:

f(1)=a
f(2)=b
f(3)=a
f(4)=b
f(5)=a
f(6)=b
f(7)=a
f(8)=b
f(9)=a
f(10)=b

both sets have more than 4 objects in. The function is not onto (no element maps to c, d, or e). The function is not one to one (note that f(1) =a and also f(3)=a)
• Mar 27th 2013, 10:25 PM
rhymin
Re: Functions / Sets / Finite Sets
Quote:

Originally Posted by Ant
Let A = {1,2,3,4,5,6,7,8,9,10}
let B = {a,b,c,d,e}

consider the map given explicitly by:

f(1)=a
f(2)=b
f(3)=a
f(4)=b
f(5)=a
f(6)=b
f(7)=a
f(8)=b
f(9)=a
f(10)=b

both sets have more than 4 objects in. The function is not onto (no element maps to c, d, or e). The function is not one to one (note that f(1) =a and also f(3)=a)

Ahh, so this would be a good example of showing that the function is not onto or one to one.

I see now that the Onto function f from A to B is called onto if for all b in B there is an a in A, such that f(a) = b. All elements in B are used.
And the one-to-one function is a function f from A to B is called one-to-one if whenever f(a) = f(b) then a = b. No element of B is the image of more than one element in A.

The illustrations from this website actually helped more than those definitions:
One-to-one and Onto Functions
• Mar 27th 2013, 10:33 PM
rhymin
Re: Functions / Sets / Finite Sets
Quote:

Originally Posted by Plato
2) $|B|^{|A|}$ is the number of functions from A to B.

Would |A| = 7 in this case?
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