I am getting frustrated with this question ( Real analysis)

[Scigatt]

office_shredder I would like to see your proof or understand what you were discribing to me.

I think that this was the proof that office_shredder was hinting at. Hopefully there aren't any errors or gaps here:

Let S be a subset of [0,1]. Consider a function f_{S}:[0,1] \rightarrow \{0,1\} such that x \in [0,1] implies f_{S}(x) = 1\ iff\ x \in S. If a subset S has two such functions, then either they are equal or there exists an x that is both an element and a non-element of S, which is impossible, so for each subset of [0,1] there exists a unique function f_S. Also, if a function f_T is the function for both T and T'(T, T' being subsets of [0,1]), then if f_T(x) = 1, then x is in both T and T' and if f_T(x) =/= 1, then x is not in either T or T', thus T and T' have exactly the same elements, and T = T'. Thus for each such function there exists a single subset of S. Also, the set of all functions with with codomain {0,1} that are related to subsets of [0,1] is itself a subset of A. Thus there is a one to one correspondence between the set of subsets of [0,1](i.e. the power set of [0,1]) and the subset of functions in A that can be defined as stated in the start of this proof. Thus, you cannot have an onto function from [0,1] to this subset of A, never mind all of A.

 

[╔(σ_σ)╝]

I think that this was the proof that office_shredder was hinting at. Hopefully there aren't any errors or gaps here:

Let S be a subset of [0,1]. Consider a function f_{S}:[0,1] \rightarrow \{0,1\} such that x \in [0,1] implies f_{S}(x) = 1\ iff\ x \in S. If a subset S has two such functions, then either they are equal or there exists an x that is both an element and a non-element of S, which is impossible, so for each subset of [0,1] there exists a unique function f_S. Also, if a function f_T is the function for both T and T'(T, T' being subsets of [0,1]), then if f_T(x) = 1, then x is in both T and T' and if f_T(x) =/= 1, then x is not in either T or T', thus T and T' have exactly the same elements, and T = T'. Thus for each such function there exists a single subset of S. Also, the set of all functions with with codomain {0,1} that are related to subsets of [0,1] is itself a subset of A. Thus there is a one to one correspondence between the set of subsets of [0,1](i.e. the power set of [0,1]) and the subset of functions in A that can be defined as stated in the start of this proof. Thus, you cannot have an onto function from [0,1] to this subset of A, never mind all of A.

Wow *faints*.


Although the proof petek help me cool up with is much shorter this is such a cool proof! Regardless of the amount of hints Office_shredder gave, I would not have come up with this argument.

I love PF simply because of this kind of things; people have such cool and different ways of doing things and this is fun to watch.


I guess proofs like this just come with maturity. Thanks a lot for the proof.

 

[Petek]

Yeah it should be a not equal sign; sorry about that. I was typing hurriedly from my phone
So I guess we are finally done.

I greatly appreciate all your help and your patience.

I will like to ask again, what do you think is the level of difficulty of this problem. I can't help but think that I sent way more time than I should have on the question.
And also is this the same proof you had in mind or was yours a little different ? If it was, I would like to see it.

The solution I had in mind was essentially the same one that you came up with. I would have defined g like this:

g(\alpha) = f_\alpha(\alpha) + 1

That makes the definition more explicit, but that's a minor point. I can't say how difficult the problem is. It depends on lots of factors. If this problem was assigned to a class, you might want to ask your fellow students about it. Or you could ask the instructor what percentage came up with the right solution.

 

[╔(σ_σ)╝]

The solution I had in mind was essentially the same one that you came up with. I would have defined g like this:

g(\alpha) = f_\alpha(\alpha) + 1

That makes the definition more explicit, but that's a minor point. I can't say how difficult the problem is. It depends on lots of factors. If this problem was assigned to a class, you might want to ask your fellow students about it. Or you could ask the instructor what percentage came up with the right solution.

Alright. Again, thank you very much for your help! I really appreciate it :-).

 

[FallMonkey]

I think that this was the proof that office_shredder was hinting at. Hopefully there aren't any errors or gaps here:

Let S be a subset of [0,1]. Consider a function f_{S}:[0,1] \rightarrow \{0,1\} such that x \in [0,1] implies f_{S}(x) = 1\ iff\ x \in S. If a subset S has two such functions, then either they are equal or there exists an x that is both an element and a non-element of S, which is impossible, so for each subset of [0,1] there exists a unique function f_S. Also, if a function f_T is the function for both T and T'(T, T' being subsets of [0,1]), then if f_T(x) = 1, then x is in both T and T' and if f_T(x) =/= 1, then x is not in either T or T', thus T and T' have exactly the same elements, and T = T'. Thus for each such function there exists a single subset of S. Also, the set of all functions with with codomain {0,1} that are related to subsets of [0,1] is itself a subset of A. Thus there is a one to one correspondence between the set of subsets of [0,1](i.e. the power set of [0,1]) and the subset of functions in A that can be defined as stated in the start of this proof. Thus, you cannot have an onto function from [0,1] to this subset of A, never mind all of A.

Now that Fs is function from [0,1] to {0,1}, then how could f(x)=1 if 1 is not in the codomain? And what do you mean by " functions in A that can be defined as stated in the start of this proof", if we only define it from [0,1] to {0,1}?

Am I missing something? Though this is a classic proof and it should hold.

 

[╔(σ_σ)╝]

Now that Fs is function from [0,1] to {0,1}, then how could f(x)=1 if 1 is not in the codomain? And what do you mean by " functions in A that can be defined as stated in the start of this proof", if we only define it from [0,1] to {0,1}?

Am I missing something? Though this is a classic proof and it should hold.

The proof is correct.

The function is well defined. For x in the subset S , of [0,1] , f(x)=1 and if x is not in S f(x) is 0. The range of f has two elements 0 and 1 so I don't know what you mean by 1 is not in the codomain. {0,1} has elements 0 and 1.

What was accomplised in the proof was that a injective was shown between the power set of the unit interval to the set all functions with range in {0,1}.
These functions are real valued and defined on the unit interval. So the means the cardinality of A is at least the cardinality of the power set of the unit interval.

The power set of the unit interval does not have the same cardinality with the unit interval thus, there is no map from [0,1] to A.

I don't see anything wrong with the proof.

 

[FallMonkey]

The proof is correct.
The function is well defined. For x in the subset S, of [0,1], f(x)=1. The range of f has two elements 0 and 1 so I don't know what you mean by 1 is not in the codomain. {0,1} has elements 0 and 1.

What was accomplised in the proof was that a injective was shown between the power set of the unit interval to the set all functions with range in {0,1}.
These functions are real valued and defined on the unit interval. So the means the cardinality of A is at least the cardinality of the power set of the unit interval.

The power set of the unit interval does not have the same cardinality with the unit interval thus, there is no map from [0,1] to A.

I don't see anything wrong with the proof.

Ahh, my stupidity. I mistook {1,0} for (0,1). Sorry and thanks for your kindness.

Now everything explains.

 

[╔(σ_σ)╝]

It's okay. Everyone makes mistakes :-).