Proving that the set of all real-valued functions is infinite dimensional without the typical infinite dimensional subspace argument The proof that the set of all real-valued functions on $\mathbb{R}$, denoted $F(\mathbb{R})$, is infinite dimensional usually follows from showing that it has an infinite dimensional subspace (in particular, the space of all polynomials on $\mathbb{R}$ -- incidentally, does this argument rely on deeper results from algebra, that say $x^{n+1}$ cannot be written as a linear combination of polynomials of degree $n$ or lower?). I'm trying to prove this simple statement without appealing to that (just because of how the textbook I'm reading presents things) and I can't seem to do it.
As (I think) is standard, the textbook defines a vector space as being finite dimensional if it equals the span of some finite set. I therefore sought to prove the result as follows:

*

*Let $A = \{{f_1,...,f_n}\}$ be a finite subset of $F(\mathbb{R})$, and suppose for the sake of contradiction that every $f \in F(\mathbb{R})$ can be written as a linear combination of these elements of $A$.

*Construct a $f$ such that this is not true.

I just can't come up with the appropriate construction. From prior experience I know it should be some variant on an arbitrary linear combo of elements of $A$, but I just can't quite think of one.
 A: I can attempt to find a solution along your strategy like this: suppose you have $f_1, \dots, f_n$ a familly of real functions on $\mathbb{R}$ and denote $V$ the span of these vectors. Then I will look at values of functions of $V$ on the integers $0, \dots, n$.
The application $G:V \rightarrow \mathbb{R}^{n+1}$ defined by $G(f) = (f(0), \dots, f(n))$ is linear. $V$ is a finite dimensional space of dimension smaller than $n$ (because it is spanned by $f_1, \dots f_n$). So $G(V)$ is a linear subspace of $\mathbb{R}^{n+1}$ of dimension smaller than $n$, in particular it is not the whole space. This means that it exists $(a_0, \dots, a_n) \in \mathbb{R}^{n+1}$ such that no function of $V$ takes exactly the values $f(i) = a_i$ for every integer $i$ between $0$ and $n$.
So to find a function $f$ not in the span $V$ of $f_1, \dots, f_n$, you just have to find a function on $\mathbb{R}$ that takes precisely the values $a_i$ on $i$, and I think you can imagine how to do that.
A: Hint: the vector space $F(\Bbb{R}) = \Bbb{R} \to \Bbb{R}$ is isomorphic to its subspace $$P = \{f : \Bbb{R} \to \Bbb{R} \mid \forall x \le 0.f(x) = 0\}$$ using your favourite bijection between $\Bbb{R}_{>0}$ and $\Bbb{R}$ (e.g. the logarithm function) to translate functions on all reals to functions on positive reals. Clearly $P$ is a proper subspace of $F$, but no finite-dimensional vector space is isomorphic to a proper subspace of itself, so $F(\Bbb{R})$ cannot be finite-dimensional. I leave it to you to translate this into a construction of the sort that you are looking for.
There is nothing special about the reals here: if $X$ is any infinite set and $\Bbb{F}$ is any field, using an injection of $X$ into a proper subset $Y$ of $X$ that is equipollent to $X$, we get an isomorphism of $X \to \Bbb{F}$ with $Y \to \Bbb{F}$ showing that $X \to \Bbb{F}$ is not finite-dimensional.
A: We can do this via diagonalization.
Suppose we have functions $f_1,...,f_n:\mathbb{R}\rightarrow\mathbb{R}$. Fix some bijection $b:\mathbb{R}\rightarrow\mathbb{R}^n$, and consider the following function:
$$g(s)=1+b(s)\cdot\langle f_1(s),...,f_n(s)\rangle.$$
(here I'm thinking of $b(s)$ as a vector, and "$\cdot$" is the dot product.)
For any tuple $r_1,...,r_n$ of reals, we have $$g\not=\sum_{i=1}^nr_if_i$$ since in particular we have $$g(b^{-1}(r_1,...,r_n))\not=\sum_{i=1}^nr_if_i(b^{-1}(r_1,...,r_n)).$$ Note that this inequality is guaranteed by the "$1+$" in the definition of $g$; think by analogy of the standard diagonal argument.
It's separately worth noting that this argument also shows that $F(\mathbb{R})$ has uncountable dimension: since $\vert\mathbb{R}^\mathbb{N}\vert=\vert\mathbb{R}\vert$ so we can run the same argument. Moreover, everything here is completely constructive, in the sense that given a finite (or countable) subset of $F(\mathbb{R})$ we explicitly produce an element of $F(\mathbb{R})$ not in the span of that set.
