Show that the vector space of real functions is not finitely spanned

Question

Here's what I've come up with. Let $\mathcal B$ be a generating set of $\mathcal F(R)$. Let us assume that $\mathcal B$ is a finite set, thus: $\lt \mathcal B \gt = \lt \{b_1, ..., b_n \}\gt, n \in N$. Let $\mathcal f \in F(R).$ Thus: $\mathcal f = \alpha_1 b_1 + ... + \alpha_nb_n,$ where $ \alpha_1,..., \alpha_n \in R$ are not all zeroes.

If $\mathcal F(R)$ is a vector space, so there can be a function $\mathcal g \in F(R)$ such as $\mathcal g \neq \lambda f, \lambda \in R$. Thus $\mathcal f+g \in F(R)$ but $\mathcal f+g \neq \alpha_1 b_1 + ... + \alpha_nb_n$. If is $\mathcal g$ is to belong to $ \lt \mathcal B \gt$ we must add a new vector to $\mathcal B$.

Repeating the above mentioned procedure, we can find a function $\mathcal g\in F(R)$ such as $\mathcal h \neq \mathcal\lambda g$. Thus, we can find infinite functions and thus need infinite base vectors.

Is this proof right?

Answer 1

I think I understand the spirit of your proof, but it doesn't seem rigorous.

What you are trying to prove is equivalent to saying that $\mathcal{F}(R)$ is an infinite-dimensional vector space (if $\mathcal{F}(R)$ were finite dimensional, any basis would be a finite spanning set). Therefore if we could display an infinite collection of linearly independent vectors, we'd be done.

Can you show that the set $\{1, x, x^2, \dots \}$ is linearly independent?

Answer 2

Here is a pretty silly proof:

Let $S$ be any infinite set, and let $F(S, \mathbb{R})$ be the $\mathbb{R}$-vector space of functions $S \to \mathbb{R}$. Furthermore, let $\mathbb{R}_d [x]$ denote the $\mathbb{R}$-vector space of polynomials in $x$ of degree $\le d$.

Suppose $f_1, \cdots, f_n \in F(S, \mathbb{R})$ is a finite basis.

Choose some $f \in F(S, \mathbb{R})$ that takes on infinitely many values. For each polynomial $p \in \mathbb{R}_d [x]$, we may write $p(f)$ uniquely in the form $\sum_{i=1}^{n} a_i f_i$ for some real scalars $a_i$. Hence, we get a linear map $\mathbb{R}_d [x] \to \mathbb{R}^n$ given by $$p \mapsto (a_1, \cdots, a_n).$$ In fact, this map is injective, since $p(f) = 0$ implies $p = 0$, given as a polynomial may have only finitely many roots.

Thus, $\dim(\mathbb{R}_d [x]) \le n$ for all $d$; taking $d\to\infty$ gives a contradiction.