5
$\begingroup$

I'm working my way through Axler's "Linear Algebra Done Right" (3rd ed.), and I'm getting stuck on section 1.23, which says:

  • If $S$ is a set, then $\textbf{F}^S$ denotes the set of functions from $S$ to $\textbf{F}$.
  • For $f, g \in \textbf{F}^S$, the sum $f + g \in \textbf{F}^S$ is the function defined by $$(f + g)(x) = f(x) + g(x)$$ for all $x \in S$.
  • For $\lambda \in \textbf{F}$ and $f \in \textbf{F}^S$, the product $\lambda f \in \textbf{F}^S$ is the function defined by $$(\lambda f)(x) = \lambda f(x)$$ for all $x \in S$.

As an example of the notation above, if $S$ is the interval [0,1] and $\textbf{F} = \textbf{R}$, then $\textbf{R}^{[0,1]}$ is the set of real-valued functions on the interval [0,1].

In the next paragraph, the author goes on to assert the following:

Our previous examples of vector spaces, $\textbf{F}^n$ and $\textbf{F}^\infty$, are special cases of the vector space $\textbf{F}^S$ because a list of length $n$ of numbers in $\textbf{F}$ can be thought of as a function from {1, 2, ..., $n$} to $\textbf{F}$ and a sequence of numbers in $\textbf{F}$ can be thought of as a function from the set of positive integers to $\textbf{F}$. In other words, we can think of $\textbf{F}^n$ as $\textbf{F}^{\{1,2,...,n\}}$ and we can think of $\textbf{F}^\infty$ as $\textbf{F}^{\{1,2,...\}}$.

It's at this point I get confused, due mostly to the example which relies on $\textbf{R}^{[0,1]}$. It seems to me that then number of elements in $\textbf{R}^{[0,1]}$ should be uncountably infinite and that I should be able to generate any value between $-\infty$ and $\infty$ from [0,1] using some member of $\textbf{R}^{[0,1]}$, which feels a whole lot like generating a point in $\textbf{R}^\infty$.

If that's the case, then what's the difference between a tuple generated from $\textbf{R}^{\{1,2,...,n\}}$ and one generated from $\textbf{R}^{\{1,2,...,n+1\}}$?

I think my difficulty lies in not understanding implicit restrictions on the notation. The answer to this specific sub-question may be a shortcut to understanding: Suppose I'm trying to think of a particular point in $(x,y,z)\epsilon\textbf{R}^3$ as $\textbf{R}^{\{1,2,3\}}$ where $f,g,h\epsilon\textbf{R}^{\{1,2,3\}}$. Must I think of that point in $\textbf{R}^3$ as $(f(1)=x, f(2)=y, f(3)=z)$, or can I think of it as $(f(1)=x, g(2)=y, h(3)=z)$?

$\endgroup$
1
  • $\begingroup$ $(x,y,z)\in\Bbb R^3$ corresponds to the function $f\in \Bbb R^{\{1,2,3\}}$ with $f(1)=x,f(2)=y,f(3)=z$. In other words, a single point in $\Bbb R^3$ corresponds to a single function in $\Bbb R^{\{1,2,3\}}$. $\endgroup$
    – Arthur
    Commented Jul 3, 2016 at 16:05

1 Answer 1

4
$\begingroup$

It's the former. The point normally written $(2, -3, 5)$ corresponds, in this book's description, the function $f: \{1, 2, 3\} \to \mathbb R$ with $f(1) = 2, f(2) = -3,$ and $f(3) = 5$.

$\endgroup$
2
  • $\begingroup$ This makes sense for sets with discreet, finite numbers of elements (i.e. the number of elements in $S$ for $\mathbb{F}^S$ corresponds to the dimensionality of the vector space), but how does this work for a continuous interval? $\endgroup$ Commented Jul 3, 2016 at 16:46
  • $\begingroup$ Well, an instance of an element of $\mathbb R^{[0, 1]}$ would be the function $g:[0, 1] \to \mathbb R: x \mapsto 2x$. In the case of this space, the "number of elements" of $[0, 1]$ also corresponds to a kind of dimensionality of the vector space, although it's rather harder to define "dimension" in this case than in the finite dimensional space. (In particular, the notion of "linear combination" where there are uncountably many things being summed is at best problematic.) $\endgroup$ Commented Jul 3, 2016 at 18:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .