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Is there a way to define a vector space of continuous functions of type $[0,1] \to [0,1]$ over $\mathbb R$? If no, how to prove that such a vector space does not exist?

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up vote 7 down vote accepted

Certainly functions with a restricted (but nonzero) range never naturally form an $\def\R{\Bbb R}\R$-vector space: if you want scalar multiplication to act by multiplying the values functions (in each point), then you can just take any point with nonzero value, and scalar-multiply it to have a value outside the specified range.

If you don't require the operations to be the ordinary ones, then defining a vector space structure on some set $X$ can be done by defining a bijection (nothing more than that) $\phi$ to a known vector space $V$. Addition in $X$ will then be done by defining $x+y=\phi^{-1}(\phi(x)+\phi(y))$ and scalar multiplication by $\lambda x=\phi^{-1}(\lambda\phi(x))$; you can verify that checking the axioms in $X$ now boils down to checking the axioms in $V$, where they hold.

Once the vector space structure on $X$ is so defined, $\phi$ actually becomes an isomorphism $X\to V$.

So if you are free to choose the operations, you are asking if the continuous functions $[0,1]\to[0,1]$ are in bijection with any $\Bbb R$-vector space. Since a continuous function is determined by its values at rational points, the cardinality of your set of function is at most that of the set of sequences of real numbers, of equivalently of $\R[[X]]$, but that cardinality turns out to be no more than that of$~\R$, so we are talking of a set with the same cardinality as any one of $\R$, $\R^n$, $\R[X]$ or $\R[[X]]$; that gives us plenty of choice for a vector space to be in bijection with. Obviously these are not going to be nice bijections, and your vector space operations will be correspondingly weird.

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If only the codomain were $(0,1)$, then we'd have nice bijections with the continuous functions $[0,1] \rightarrow (-\infty, \infty)$. (Where $ (-\infty, \infty)$ is a suggestive way of writing $\mathbb{R}$) – Steve Jessop Aug 31 '14 at 18:27
@SteveJessop: That's right. And since an explicit (though non-continuous) bijection $[0,1]\to(0,1)$ can be given, this allows us to define an explicit vector space structure on the given set of functions; not really a nice one, but at least something that does not depend on the axiom of choice, as the argument in my answer does. – Marc van Leeuwen Sep 1 '14 at 9:34

Call $C = \{ f: [0,1] \longrightarrow [0,1] \mbox{ continuous functions } \}$. This is a set.

You are asking if there exists any way to define on it a structure of real vector space. Well, the answer is: if you don't require anything else, then yes, there exists a way. But this is a stupid way of doing it.

Simply consider any real valued vector space $V$ with the same cardinality of $C$. Then there is a bijection $l: C \longrightarrow V$. I call this bijection $l$ like '$label$'.

Then define for $f,g \in C$ and $\lambda \in \mathbb{R}$ $$f+g = l^{-1}(l(f) + l(g))$$

$$ \lambda f = l^{-1} (\lambda l(f))$$

With this formal definition of sum and scalar multiplication, you endow $C$ with a structure of a real-valued vector space, just isomorphic to $V$.

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Consider any non-identically-zero function $f : [0,1] \longrightarrow [0,1]$. Then $-f$ is not a function $[0,1] \longrightarrow [0,1]$, so your set is not a group, hence neither a vector space.

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And if we try to define the operations differently? – Amitai Yuval Aug 31 '14 at 11:26
Do you assume (-f)(x) = -f(x)? If so, we have to prove there is no other way to define inverse (and other stuff) to make it a vector space. – aemxdp Aug 31 '14 at 11:33
Well, I don't think there exists any natural way to define an abelian group with this set. What you can always do is to identify it with any vector space of the same cardinality with any bijection, but this is not natural. – Crostul Aug 31 '14 at 11:38

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