Wednesday, October 28, 2009

New questions

1. Let C[a,b] be the set real-valued continuous functions on [a,b], and Let Cst[a,b] be the set of all constants functions on [a,b]. What is the cardinality of C[a,b]/Cst[a,b]?
2. Does C[a,b] have a countable Schauder basis?
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5 comments:

  1. UnknownOctober 29, 2009 at 5:43 PM

    For Question 1:

    The cardinality of the C[a,b] is equal to c since [a,b] is separable. It seems reasonable that cardinality of C[a,b]/Cst[a,b] is also c. But I can't really think of a homomorphism that has kernel Cst[a,b].

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  2. MathblogOctober 29, 2009 at 11:02 PM

    Jim,
    Let D[a,b] be the set of all differentiable functions over [a,b]. Consider the linear map F defined on D[a,b] such that F(f)=f'. KerF=Cst[a,b]. Thus D[a,b]/Cst[a,b] is isomorphic to F(D[a,b]) which is the set of all functions that have continuous antiderivatives. Since D[a,b] is included in C[a,b], then C[a,b]/Cst[a,b] is included in D[a,b]/Cst[a,b]. Thus, C[a,b]/Cst[a,b] is smaller than some isomorphic copy of F(D[a,b]). I am not sure how big that space is. Does anyone know how big is the space of all functions that have continuous antiderivatives?

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  3. UnknownOctober 30, 2009 at 10:54 PM

    Is it true that cardinality of the Cst[a,b] is also c? I don't know. I do know that the cardinality of C[a,b] = c because [a,b] is separable so there is a countable dense subset namely Q \cap [a,b]. Since a continuous function is determined on a dense set, we can have f: C[a,b] --> Q \cap [a,b] such that f=f_0 restricted to Q \cap [a,b]. So, then we have that f is 1-1, thus |C[a,b]| =< |R^Q| = c^\aleph_0 = c. Okay, but now I am stuck with this quotient thing...

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  4. Vignon OussaOctober 31, 2009 at 8:56 PM

    Jim, I am not sure why the fact that [a,b] is seperable makes the cardinality of C[a,b]the continium. All we can say about a separable space is that it has a cardinality of at most 2^c.I think you want to show that C[a,b] contains a countable dense set. The set of polynomial is dense when C[a,b] is endowed with the uniform norm topology. So that makes C[a,b] separable. However, it still doesn't answer the question.

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  5. UnknownOctober 31, 2009 at 11:18 PM

    Separability of [a,b] is needed in the proof to extract a dense subset namely Q \cap [a,b].

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