Help with proof, characteristic function of the subset

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Homework Help Overview

The discussion revolves around proving the cardinality of the set of functions mapping a set A into the set B={0,1}, denoted as B^{A}, and showing that it is equal to the cardinality of the power set of A. Participants are exploring the relationship between functions and subsets in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the concept of cardinality and the nature of functions from A to B. There are attempts to clarify the proof structure, with some questioning how to explicitly define a mapping between subsets of A and functions in B^{A}. Others express uncertainty about the definitions and implications of the characteristic function.

Discussion Status

The discussion is ongoing, with participants providing insights and seeking clarification on the mapping between subsets and functions. Some have suggested a potential approach involving characteristic functions, while others are still grappling with the definitions and the proof structure.

Contextual Notes

Participants are navigating the definitions of functions and subsets, as well as the requirements for establishing a one-to-one mapping. There is a noted lack of consensus on the clarity of the proof and the explicit definitions needed for the mappings involved.

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Homework Statement



You have a set A. Let [tex]B^{A}[/tex] be the set of all functions mapping A into the set B={0,1}. Prove the cardinality of[tex]B^{A}[/tex] = the cardinality of the power set of A.

2. Homework Equations [/b

The Attempt at a Solution


I feel about 50% sure about the proof. What do you think? Since A is a set, A has n elements where n is a natural number. Let g be the function mapping A into B s.t {n[tex]\in[/tex]A : n = 0 or n = 1}. So for each n in A there are 2 distinct possibilities (i. e mapping to 0 or 1). Since [tex]B^{A}[/tex] is the set containing all functions of type g, the cardinality of [tex]B^{A}[/tex] = [tex]2^{n}[/tex], which is also known as the cardinality of the power set of A.

Thanks for helping!
 
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If you want to show the cardinalities are equal the easiest way is to exhibit a 1-1 mapping between subsets of A and members {0,1}^A. Your 'proof' doesn't actually ever say what it is. Can you be more explicit? If you have subset S of A define the mapping. If you have a mapping define the subset.
 
I'm not sure how to create a mapping from the power set of A into the interval? Since the problem talks about the functions of A, I'm not sure what the power set of A would look like?
 
It's not an interval. B={0,1} is just the set consisting of the elements 0 and 1. B^A is just the set of all mappings from A->B, i.e. all functions f, such that f(a)=0 or 1 for each a in A. If you have a subset C of A what's the obvious way to define the corresponding function f? If you have a function f what's the corresponding subset C?
 
You said there is an obvious function, but it is not obvious to me. Would the function C (the subset of A) going to B be the function of P(A)? I know we have to show there is a 1-1 function from P(A) to B (to the A). I have no idea how.
 
Given a subset C, to me the obvious function f is f(a)=0 if a is not in C and f(a)=1 if a is in C. Given a function f define C={a such that f(a)=1}. It's called the characteristic function of the subset.
 

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