I am getting frustrated with this question ( Real analysis)

[╔(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 :-).