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Space of continuous functions C[a,b]

  1. Jul 22, 2010 #1
    We know that dim(C[a,b]) is infinte. Indeed it cannot be finite since it contains the set of all polynomials.
    Is the dimension of a Hamel basis for it countable or uncountable?

    I guess if we put a norm on it to make a Banach space, we could use Baire's to imply uncountable.
    I am however interested in a proof that does not rely on equipping the vector space with a norm.

    If it's too long winded but available somewhere else(books, articles) kindly point it out to me and I'll read it.
  2. jcsd
  3. Jul 22, 2010 #2
    Hamel bases are always finite or uncountable.
  4. Jul 22, 2010 #3


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    That's true if your space is an infinite dimensional normed vector space which is complete, and that result follows from Baire's category theorem AFAIK. So you haven't really strayed from the original proof.

    A counting argument might suffice. Suppose there was a countable basis. There are only continuum many choices of finite linear combinations from these, but the cardinality of the vector space is larger than that
  5. Jul 26, 2010 #4
    I used instead the fact that exponentials are continuous
    Then E:={e^ax : a real } is a subset of C[a,b]

    It's clear that for finite n, {e^ix : i in N} is linearly independent.
    I'm fudging, but that would seem to imply that E is also linearly independent, since every finite subset is linearly independent.
    Hence cardinality of a basis is at least |E| = continuum
  6. Jul 31, 2010 #5
    False: the cardinality of C[a,b] is the cardinality of the continuum. (Each function in C[a,b] is uniquely determined by its values on the rational numbers in [a,b].)

    But yes, you could show that E = {eax | a real} is an uncountable linearly independent subset of C[a,b], to show that C[a,b] has no countable Hamel basis.

    The space of all real sequences which are eventually zero has a countable Hamel basis.
    Last edited: Jul 31, 2010
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