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While reading a book Linear Algebra Done Right, I came to knew that a vector space $\mathbf{R}^n$ represents a space with dimensions as $(x_1, x_2, ...,x_n)$, but there were other vector spaces that I could not understand.

There was a statement as Ref: 1.35

The set of continuous real-valued functions on the interval $[0,1]$ is a subspace of $\mathbf{R}^{[0,1]}$

What kind of space does $\mathbf{R}^{[0,1]}$ represent? Is this a space that can continuously be from $0$ dimension to $1$ dimension?

Another statement, Ref: 1.35

The set of differentiable real-valued functions on $\mathbf{R}$ is a subspace of $\mathbf{R}^\mathbf{R}$

What kind of space is $\mathbf{R}^\mathbf{R}$?

Similarly, there were other subspaces as, $\mathbf{R}^{(0,\ 3)}$ and $\mathbf{R}^{(-4,\ 4)}$

Explain me how can I visualize such spaces. If you can explain with the proof too, that will be great.

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When $A$ and $B$ are sets, $A^B$ represents a set. The elements of $A^B$ are functions from $B$ to $A$.

A vector of $\mathbb R^{[0,1]}$ is a function that maps $[0,1]$ to the real numbers. For example, $$f:[0,1]\to \mathbb R\\ f(x)=x^2$$ is one such vector.

Addition and scalar multiplication in the vector space are defined as

  • If $f, g\in\mathbb R^{[0,1]}$, then $f+g$ is defined as $(f+g)(x)=f(x)+g(x)$
  • If $f\in\mathbb R^{[0,1]}$ and $\lambda\in\mathbb R$ then $\lambda f$ is defined as $(\lambda f)(x)=\lambda\cdot f(x)$

It's easy to see that the functions $f+g$ and $\lambda f$ above are both also elements of $\mathbb R^{[0,1]}$

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  • $\begingroup$ @Michael Yeah. Thanks. $\endgroup$ – 5xum Jan 8 '18 at 13:50
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The set $\mathbb{R}^{[0,1]}$ is the set of all functions from $[0,1]$ into $\mathbb R$ and the set $\mathbb{R}^\mathbb{R}$ is the set of all functions from $\mathbb R$ into itself.

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There's nothing wrong with the other answers, but just to tie together some ideas:

In an $n$ dimensional real vector space, the vectors are tuples $(x_1, ..., x_n)$. You can think of these tuples as functions $f: \{ 1,...n \} \to \mathbb{R}$. Functions can be added and scalar multiplied pointwise, as 5xum points out, and so really the uncountable case, where the 'tuple' is instead a function from $[0,1] \to \mathbb{R}$ is exactly the same as the situation you're already familiar with, but with infinitely many numbers in the 'tuple'.

It's worth noting that there is a vector space isomorphism between $\mathbb{R}^X$ and $\mathbb{R}^Y$ for any two uncountable sets $X \to Y$ with the same cardinality, since they differ only by changing the way you refer to the basis elements by using a set bijection. Thus in particular as vector spaces $\mathbb{R}^{[0,1]} \simeq \mathbb{R}^{\mathbb{R}}$

Finally, if you want a way to visualize these guys, you can actually think about them like a graph of a function (if the function is not so badly discontinuous this gets impossible). The reason is because each slice of the graph can be thought of as one of the uncountably many values the function takes. For every real number in $[0,1]$, this function chooses another real number, and so you can imagine putting all these values 'side by side' to get the graph. For nice functions this can make it easy to see what a elements of this vector space may look like, but be aware that unless you're dealing with the vector sub-space of continuous functions, most vectors are going to be pretty badly discontinuous as functions. But at least this view makes it easy to see that functions do form a vector space!

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  • $\begingroup$ "It's worth noting that there is a vector space isomorphism between $\mathbb{R}^X$ and $\mathbb{R}^Y$ for any two uncountable sets $X \to Y$, since they differ only by changing the way you refer to the basis elements by using a set bijection." But $X$ and $Y$ may have different cardinality, say $X=\mathbb{R}$, $Y=2^\mathbb{R}$. $\endgroup$ – JiK Jan 8 '18 at 14:24
  • $\begingroup$ @JiK Ah, thanks. Early in the morning. I'll edit that. $\endgroup$ – Alfred Yerger Jan 8 '18 at 14:40

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