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.

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- June 26th 2011, 05:47 AMmathabca 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. - July 17th 2011, 06:00 PMojonesRe: a question about the cantor set
Did you get an answer to this?

- July 17th 2011, 11:25 PMmathabcRe: a question about the cantor set
- July 18th 2011, 11:07 AMHallsofIvyRe: 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."

- July 19th 2011, 04:21 PMTinybossRe: a question about the cantor set
- July 19th 2011, 06:21 PMojonesRe: a question about the cantor set
- July 19th 2011, 06:24 PMTinybossRe: a question about the cantor set
- July 19th 2011, 07:30 PMojonesRe: a question about the cantor set
- July 20th 2011, 01:22 PMOpalgRe: a question about the cantor set
We are told nothing about these sets , 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 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 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 then at least one of those sets must have uncountable intersection with uncountably many equivalence classes, say is uncountable for all [x] in an uncountable set X of equivalence classes. I now want to say that there must be a point that is a limit point for at least two distinct sets and where [x] and [y] are in X. It will then follow that the restriction of f to is discontinuous at z (because z is the limit of two sequences in , 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. - July 20th 2011, 11:45 PMJose27Re: 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) - July 21st 2011, 12:56 AMOpalgRe: a question about the cantor set
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.

- July 21st 2011, 04:51 PMojonesRe: a question about the cantor set