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While the elements of $[0,1]^{\mathbb{R}}$ satisfy properties of vector spaces (commutativity, associativity etc..), I get the feeling that you can find $f,g \in [0,1]^{\mathbb{R}}$ such that for some $x \in \mathbb{R}$ you have $f(x) + g(x) \not \in [0,1]$.

The reason I'm struggling is that in Linear Algebra done right (page 14), we prove that

If S is a set, and $\mathbb{F}^S$ denotes the set of functions from S to $\mathbb{F}$, then $\mathbb{F}^S$ is a vector space.

Where am I going wrong?

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    $\begingroup$ Are you talking about $[0,1]^{\mathbb{R}}$ or ${\mathbb{R}}^{[0,1]}$? $\endgroup$ Commented May 7, 2022 at 16:16
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    $\begingroup$ If $f(x)=1$ and $g(x)=1$ enough to see that it is not a vector space. I think you want to study $\mathbb{R}^{[0,1]}$ $\endgroup$ Commented May 7, 2022 at 16:18
  • $\begingroup$ If you're talking about the set of functions from $[0,1]$ to $\mathbb{R}$, you'd be choosing $x\in[0, 1]$ and your images $f(x)$ would live in $\mathbb{R}$. $\endgroup$ Commented May 7, 2022 at 16:19
  • $\begingroup$ No I mean $[0,1]^{\mathbb{R}}$. Is it not generally true that if S is a set, and $\mathbb{F}^S$ denotes the set of functions from S to $\mathbb{F}$, then $\mathbb{F}^S$ is a vector space? Why doesn't that work here? $\endgroup$
    – Darby Bond
    Commented May 7, 2022 at 16:25

2 Answers 2

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When we study a vector space $V$, we have a structure algebraic $$\underbrace{(V,\overbrace{(\mathbb{F},+,\cdot)}^{\text{field}},\oplus, \odot)}_{\text{vector space}},$$where $\oplus$ and $\odot$ they're operations over $V$ and $+$ and $\cdot$ they're operations in $\mathbb{F}$.

Now, just small remarks:

  • If you don't mention the operations over $V$ and over $\mathbb{F}$ so we will assume the usual/natural operations. In that sense $([0,1],+,\cdot)$ is not a field because is not closed or just see in the axioms of fields.

  • If $S$ is set and $\mathbb{F}$ is a field and $\mathbb{F}^{S}$ denotes the set of functions from $S$ to $\mathbb{F}$, then $\mathbb{F}^{S}$ is a vector space. That's true.

  • Then under the natural operations, $([0,1]^{\mathbb{R}},([0,1],+,\cdot),\oplus, \odot)$ is not a vector space because is not closed.

  • But under the natural operations, $(\mathbb{R}^{[0,1]},(\mathbb{R},+,\cdot),\oplus,\odot)$ is a vector space just see in the axioms of vector space.

Is for that reason that is important to write all the hypothesis, it helps avoid mistakes.

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If you see the statement , it says $\Bbb{F}^{S}$ and not $S^{\Bbb{F}}$ .

Here $S=[0,1]$ and $\Bbb{F}=\Bbb{R}$.

Hence you should be looking at $\Bbb{R}^{[0,1]}$. This satisfies all properties of a vector space .

This is nothing but an uncountable cartesian product. In fact you can define a linear map $T:\Bbb{F}^{S}\to \prod_{s\in S}\Bbb{F}$ given by $T(f)=\prod_{s\in S} (f(s))$ . Then you can prove that this is an isomorphism. In fact cartesian product(can be infinite product) of vector spaces is always a vector space and is uniquely defined upto isomorphism by the universal property of products. But you can ignore the last sentence safely for now until you learn maybe tensor products and direct sums etc.

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