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Simple Indicator Function and its Set

  1. Jul 25, 2013 #1
    1. The problem statement, all variables and given/known data

    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##

    2. Relevant equations

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

    3. 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
     
    Last edited: Jul 25, 2013
  2. jcsd
  3. Jul 26, 2013 #2

    jbunniii

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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?
     
  4. Jul 26, 2013 #3
    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?
     
  5. Jul 26, 2013 #4

    jbunniii

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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.
     
  6. Jul 26, 2013 #5
    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.
     
  7. Jul 26, 2013 #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.
     
  8. Jul 26, 2013 #7
    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##.
     
  9. Jul 26, 2013 #8

    pasmith

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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?
     
  10. Jul 26, 2013 #9
    ##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?
     
    Last edited: Jul 26, 2013
  11. Jul 26, 2013 #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.
     
  12. Jul 26, 2013 #11
    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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