Simple Indicator Function and its Set

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Homework Statement



This is taken from Spivak's Calculus Book Chapter 3 - Functions, Problem 9.

Suppose ##f## is a function such that ##f(x)=1\text{ or }0## for each ##x##. Prove that there is a set ##A## such that ##f = C_A## ##C_A## is an indicator function, where ##C_A(x)=0## if ##x## is not in ##A##, and 1 if ##x## is in ##A##

Homework Equations



I've finished the previous sub-problem (a) in the same number.

The Attempt at a Solution


I don't understand clearly what does the problem wants us to prove. Is it asking us whether a set exist like this for example ##A=\{x : f(x)=1 \text{ or } 0\}## exist? How then can we prove that it is true? What kind of condition is that? I mean how can we know what lies inside of a set, with only the property, that it has to lie inside the set. It's a bit confusing..

EDIT: Isn't it like hey I tell my friend to go to a market to buy something and put it in a bucket. So what do you want to buy then? He asked. I want to buy everything that is in that bucket.

Thank You
 
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Answers and Replies

  • #2
jbunniii
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I don't understand clearly what does the problem wants us to prove. Is it asking us whether a set exist like this for example ##A=\{x : f(x)=1 \text{ or } 0\}## exist?
Well, it wouldn't be ##A=\{x : f(x)=1 \text{ or } 0\}##. Are there any values of ##x## that are NOT in that set?

You want ##A## to be the set of points ##x## where ##f(x) = ##what value?
 
  • #3
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Well, it wouldn't be ##A=\{x : f(x)=1 \text{ or } 0\}##. Are there any values of ##x## that are NOT in that set?

You want ##A## to be the set of points ##x## where ##f(x) = ##what value?
Ah I see, because it is a set of all real number ##x## that ##C_A## is referring to, then there's actually no real number which is not in the set. So ##f(x)## has to be 1. Is the reasoning right?
 
  • #4
jbunniii
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Ah I see, because it is a set of all real number ##x## that ##C_A## is referring to, then there's actually no real number which is not in the set. So ##f(x)## has to be 1. Is the reasoning right?
Right, so write out the definition of ##A## and you're pretty much done. I'm not sure why Spivak is asking for a "proof" here as there isn't much to prove.
 
  • #5
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Right, so write out the definition of ##A## and you're pretty much done. I'm not sure why Spivak is asking for a "proof" here as there isn't much to prove.
Ok sure I'll do so. On the other hand, will setting the condition of ##f(x)=0## create an empty set then?

Thanks for your help.
 
  • #6
HallsofIvy
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No, you are missing the whole point. You are told only that f(x) takes on values of 0 and 1. f(x) could be "f(x)= 0 if x is rational, 1 if x is irrational". In that case, A= {x| f(x)= 1} is the set of irrational numbers, B= {x| f(x)= 0} is the set of rational numbers.

Or you could have f(x) defined to be 0 for all x. In that case A would be empty.
 
  • #7
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No, you are missing the whole point. You are told only that f(x) takes on values of 0 and 1. f(x) could be "f(x)= 0 if x is rational, 1 if x is irrational". In that case, A= {x| f(x)= 1} is the set of irrational numbers, B= {x| f(x)= 0} is the set of rational numbers.

Or you could have f(x) defined to be 0 for all x. In that case A would be empty.
Yes it is true, but then the question says such that ##f=C_A##. While ##C_A## clearly stipulates 1 for the set of real number ##A##.
 
  • #8
pasmith
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Yes it is true, but then the question says such that ##f=C_A##. While ##C_A## clearly stipulates 1 for the set of real number ##A##.
Look at the question again:

Suppose ##f## is a function such that ##f(x)=1\text{ or }0## for each ##x##. Prove that there is a set ##A## such that ##f = C_A## ##C_A## is an indicator function, where ##C_A(x)=0## if ##x## is not in ##A##, and 1 if ##x## is in ##A##
This is asking you to prove that, for every [itex]f : \mathbb{R} \to \{0, 1\}[/itex], there exists a subset [itex]A \subset \mathbb{R}[/itex] such that [itex]f(x) = 1[/itex] if and only if [itex]x \in A[/itex] and [itex]f(x) = 0[/itex] if and only if [itex]x \in \mathbb{R} \setminus A[/itex].

There's really only one choice for A. What is it?
 
  • #9
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Look at the question again:

This is asking you to prove that, for every [itex]f : \mathbb{R} \to \{0, 1\}[/itex], there exists a subset [itex]A \subset \mathbb{R}[/itex] such that [itex]f(x) = 1[/itex] if and only if [itex]x \in A[/itex] and [itex]f(x) = 0[/itex] if and only if [itex]x \in \mathbb{R} \setminus A[/itex].

There's really only one choice for A. What is it?
##A## should be the set of real number ##\mathbb{R}## and that ##f(x)=1## because of the fact that there is an element in the set? Is there anything else in it? I'm sorry if I miss the subtleties but the language of set is a bit beyond my level right now.

Edit: Or ##A## must contain all rational number instead?
 
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  • #10
HallsofIvy
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A is NOT the set f all real numbers- it is the set of all numbers such that f(x)= 1.
 
  • #11
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A is NOT the set f all real numbers- it is the set of all numbers such that f(x)= 1.
Ok, so this function ##f(x)## so to say is general right? We don't decide for now what makes it 1? We just prepare a container for whatever might make it ##1## according to our own definition?
 

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