Can someone teach me this problem of sets.

  • Thread starter Thread starter S&S
  • Start date Start date
  • Tags Tags
    Sets
Click For Summary
SUMMARY

The discussion focuses on the relationship between functions and their mappings in set theory, specifically examining the function G defined from the power set of B to the power set of A. It establishes that G is one-to-one if and only if the function f from set A to set B is onto. Through examples using sets A = {1,2,3} and B = {a,b}, participants illustrate how different mappings affect the properties of G, demonstrating the conditions under which G can be one-to-one or onto.

PREREQUISITES
  • Understanding of set theory concepts, including functions and power sets.
  • Familiarity with the definitions of one-to-one (injective) and onto (surjective) functions.
  • Basic knowledge of examples illustrating mappings between sets.
  • Ability to analyze mathematical functions and their properties.
NEXT STEPS
  • Study the definitions and properties of injective and surjective functions in detail.
  • Explore the concept of power sets and their applications in set theory.
  • Learn about the implications of function composition in relation to one-to-one and onto properties.
  • Investigate more complex examples of functions between sets to solidify understanding of these concepts.
USEFUL FOR

Students of mathematics, particularly those studying set theory, educators teaching mathematical concepts, and anyone looking to deepen their understanding of functions and their properties in mathematical contexts.

S&S
Messages
11
Reaction score
0
Let A and B be sets and Let f:A---->B be a function. Define a function G: Power set (B)------->Power (A) by declaring that, for C is part of B or equal to B:
G(C)={x is element of A: f(x) is element of C}
Show that G is 1-1 if and only if f is onto. :cry:

Can anyone teach me this problem? Hopefully I can get some feelings about set theory. I just don't feel right.

Thank you so much for your good tips.
 
Physics news on Phys.org
When you look at a problem like this and it means nothing to you, look at simple examples. Suppose A is the set {1,2,3} and B is {a,b}.
1) First example. Define f by f(1)= a, f(2)= a, f(3)= a (f is neither 1-1 nor onto). The power set of B is {{}, {a}, {b}, {a,b}}. G({})= {}, of course, because every member of A is mapped into something. G({a})= {1, 2, 3} because every member of A is mapped into a. G({b})= {} because no member of A is mapped into b. G({a,b})= {1,2,3} because every member of A is mapped into a which is in {a,b}. This is not 1-1 because two different sets, {} and {a} are mapped into {} and, also, two different sets, {a} and {a,b} are mapped into {1, 2, 3}. It is not onto because there is no set which is mapped into, for example, {1}, a set in the powerset of A.
2) Second example. g(1)= a, g(2)= a, g(3)= b (g is onto but not 1-1).
G({})= {}, G({a})= {1,2}, G({b})= {3}, G({a,b})= {1, 2, 3}. Now G is 1-1 since no two members of the power set of B are mapped into the same thing. It is NOT onto because no set is mapped into, for example, {1, 2}.

We can't give an example with these two sets of a 1-1 function because A has too many elements. Take A= {1,2}, B= {a,b,c} and define f(1)= a, f(2)= b. Now f is 1-1 but not onto. G({})= {}, G({a})= {1}, G({b})= {2}, G({c})= {}, G({a,b})= {1,2}, G({a,c})= {1}, G({b,c})= {2}, G({a,b,c})= {1,2}. G is onto but not 1-1.

That's interesting. Perhaps we can simplify the problem by doing it as two separate parts:
(1) the function G is 1-1 if and only if f is 1-1 and
(2) the function G is onto if and only if f is onto.
 
just say cheers

I'm getting better.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 15 ·
Replies
15
Views
4K
Replies
23
Views
6K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
4
Views
4K
Replies
1
Views
3K