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.
Best Answer
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.