1. ## a question about the cantor set

Please give a real-valued function f on the cantor set C in [0,1]
satisfies
there doesn't exist a sequence of subsets {Ek}k=1,2,...
of C
such that
(1)the untion of Ek is C;
(2)the restriction of f to every set Ek is continuous on Ek.

2. ## Re: a question about the cantor set

Did you get an answer to this?

3. ## Re: a question about the cantor set

Originally Posted by ojones
Did you get an answer to this?
No,I didn't.

4. ## Re: a question about the cantor set

What about "f(x)= 1 if x is an irrational number in the Cantor set and f(x)= -1 if x is a rational number in the Cantor set."

5. ## Re: a question about the cantor set

Originally Posted by HallsofIvy
What about "f(x)= 1 if x is an irrational number in the Cantor set and f(x)= -1 if x is a rational number in the Cantor set."
That function is continuous on the inverse image of 1 and the inverse image of -1, and the union of those two sets is C. No function with countable image can satisfy the condition.

6. ## Re: a question about the cantor set

Originally Posted by Tinyboss
That function is continuous on the inverse image of 1 and the inverse image of -1, and the union of those two sets is C. No function with countable image can satisfy the condition.
This doesn't make sense; aren't you claiming that $f$ satisfies the condtions?

7. ## Re: a question about the cantor set

Originally Posted by ojones
This doesn't make sense; aren't you claiming that $f$ satisfies the condtions?
If I read the question correctly, there should not exist a sequence of subsets satisfying conditions 1 and 2. For the function Hallsofivy gave, the sequence (S,T,T,T,...) satisfies the conditions, where S and T are the inverse images of 1 and -1 respectively.

8. ## Re: a question about the cantor set

Originally Posted by Tinyboss
If I read the question correctly, there should not exist a sequence of subsets satisfying conditions 1 and 2. For the function Hallsofivy gave, the sequence (S,T,T,T,...) satisfies the conditions, where S and T are the inverse images of 1 and -1 respectively.
But you also say that "No function with countable image can satisfy the condition." I consider finite sets countable.

In any event, the statement says that no such sequence exists so something is wrong somewhere.

9. ## Re: a question about the cantor set

Originally Posted by mathabc
Please give a real-valued function f on the cantor set C in [0,1]
satisfies
there doesn't exist a sequence of subsets {Ek}k=1,2,...
of C
such that
(1)the union of Ek is C;
(2)the restriction of f to every set Ek is continuous on Ek.
We are told nothing about these sets $E_k$, so we cannot assume any topological or measure-theoretic properties for them. Therefore it seems that any construction must rely on counting and cardinality.

The Cantor set has the cardinality of the continuum, so there exists a bijective mapping $x\mapsto(f(x),\phi(x))$ from C to the unit square [0,1]x[0,1]. Thus I am defining the map f as the first coordinate of that bijection. Define an equivalence relation on C by $x\sim y \;\Leftrightarrow\; f(x)=f(y).$ There are uncountably many equivalence classes, each containing uncountably many elements of C. Denote by [x] the equivalence class containing x.

If C is the union of countably many sets $E_k$ then at least one of those sets must have uncountable intersection with uncountably many equivalence classes, say $E_k\cap[x]$ is uncountable for all [x] in an uncountable set X of equivalence classes. I now want to say that there must be a point $z\in E_k$ that is a limit point for at least two distinct sets $E_k\cap[x]$ and $E_k\cap[y],$ where [x] and [y] are in X. It will then follow that the restriction of f to $E_k$ is discontinuous at z (because z is the limit of two sequences in $E_k$, on one of which f takes the constant value f(x), and on the other it takes the constant value f(y)). I don't have the energy to write out that last part of the argument in detail, but I am more or less convinced that it works.

10. ## Re: a question about the cantor set

@Opalg: I think this question is closely related to this one, and if that's the case measure theory does indeed come strongly into play: The crucial part being if the OP would be working with Lebesgue measure or its completion, if the former then I'm not sure your argument would hold in general (at least not necessarily since there are continous non-Lebesgue measurable functions, and I don't think the Cantor set is devoid of non-measurable sets). Do you have any thoughts on this? (as you can see I haven't been able to answer said question)

11. ## Re: a question about the cantor set

Originally Posted by Jose27
@Opalg: I think this question is closely related to this one, and if that's the case measure theory does indeed come strongly into play: The crucial part being if the OP would be working with Lebesgue measure or its completion, if the former then I'm not sure your argument would hold in general (at least not necessarily since there are continous non-Lebesgue measurable functions, and I don't think the Cantor set is devoid of non-measurable sets). Do you have any thoughts on this? (as you can see I haven't been able to answer said question)
I am no longer convinced that the reasoning in my previous comment is correct, in fact the argument in the final paragraph breaks down completely. But I still have the intuitive feeling that a continuous function on an arbitrary subset E of the unit interval cannot take each of uncountably many values uncountably often. I have not been following that other thread, but I don't see how measure theory can be relevant to this current thread, since the Cantor set and all its subsets have measure 0.

12. ## Re: a question about the cantor set

Originally Posted by mathabc
Please give a real-valued function f on the cantor set C in [0,1]
satisfies
there doesn't exist a sequence of subsets {Ek}k=1,2,...
of C
such that
(1)the untion of Ek is C;
(2)the restriction of f to every set Ek is continuous on Ek.
Where did you find this problem?