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How to find the dimension of a function? I just get confused of that:

Consider the set of continuous functions on the interval $[0,1]$: $$C[0,1]=\{f:[0,1]\mapsto \mathbb{R} | f\text{ is continuous}\}$$ Then, what is the dimension of $C[0,1]$?

How can I find the dimension of $C[0,1]$ and how can I verify that?

vanguard2k
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Scorpio19891119
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1 Answers1

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$C[0,1]$ is an infinite-dimensional space. For example, $1, x, x^2, x^3 , \ldots$ are linearly independent.

Robert Israel
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  • From my intuition, it should be infinite. But how can I verify and prove that? Just give an example? I do not think that is enough. – Scorpio19891119 Sep 11 '12 at 20:45
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    You can check by defition: take $\lambda _1 ,\lambda _2, \cdots ,\lambda _n$ such that $\lambda _1 x^{k_1}+\lambda _2 x^{k_2}+\cdots+\lambda _n x^{k_n}=0 \forall x\in [0,1]$ A polynomial with infitite zeros is the trivial so you are done – clark Sep 11 '12 at 20:48
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    Also a vector space has infinite dimension if it has a set of vectors such that every finite subset of them is linearly independent – clark Sep 11 '12 at 20:50
  • A set of vectors is linearly independent if every finite subset of that set is linearly independent. That is, by definition if a set is linearly dependent it has a nontrivial linear combination that is $0$, and a linear combination involves only a finite collection of vectors. – Robert Israel Sep 11 '12 at 22:43
  • Which kind of infinite? @RobertIsrael – irene dovichi Sep 27 '20 at 15:06
  • The cardinality of a basis of $C[0,1]$, as with any separable infinite-dimensional Banach space, is $\mathfrak c$ (the cardinality of the continuum). See e.g. here. – Robert Israel Sep 27 '20 at 21:10