Simple Indicator Function and its Set

[Seydlitz]

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

 

[jbunniii]

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?

 

[Seydlitz]

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?

 

[jbunniii]

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.

 

[Seydlitz]

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.

 

[HallsofIvy]

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.

 

[Seydlitz]

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 $A$.

 

[pasmith]

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 $f : \mathbb{R} \to \{0, 1\}$, there exists a subset $A \subset \mathbb{R}$ such that $f(x) = 1$ if and only if $x \in A$ and $f(x) = 0$ if and only if $x \in \mathbb{R} \setminus A$.

There's really only one choice for A. What is it?

 

[Seydlitz]

Look at the question again:

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

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?

 

[HallsofIvy]

A is NOT the set f all real numbers- it is the set of all numbers such that f(x)= 1.

 

[Seydlitz]

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?