Does $\{1,x,x^2....\}$ span $F(\mathbb R)$, where $F(\mathbb R)$ is the set of all functions $\mathbb R \to \mathbb R$?

Question

My Approach:

I'm assuming that $\{1,x,x^2,\dots\}$ is $P(R)$, the set of all polynomials.

Suppose $\{1,x,x^2,\dots\}$ spans $F(R)$. Then by definition $F(R) = \text{span } P(R)$. However, $P(R)$ is a subspace of $F(R)$ and as a result, $P(R)$ can not span $F(R)$.

Sorry in advance if this is not the best proof/approach.

Answer 1

It would be simpler to note that, since each of the functions $1$, $x$, $x^2$, … is continuous, all the functions from the space that they span is continuous too. But there are discontinuous functions in $F(\mathbb{R})$.

Answer 2

$P(\mathbb{R})$ is a subspace of $F(\mathbb{R})$

This line of reasoning screams two things to me.

• Firstly, a linguistic problem: "subspaces" don't have to be proper subspaces: that is, the two could still be equal. (For example, $F(\mathbb{R})$ is a subspace of $F(\mathbb{R})$ - it's the biggest one there is!)
• Secondly, a logical problem. Even if we replace "subspace" by "proper subspace": you've claimed that $P(\mathbb{R})$ is a proper subspace of $F(\mathbb{R})$, but how do you know that? The phrase "$P(\mathbb{R})$ is a subspace of $F(\mathbb{R})$" is actually just a fancier way of saying "$\{1, x, x^2,\dots\}$ does not span $F(\mathbb{R})$", rather than a proof of that statement.

There are many reasons why $P(\mathbb{R})$ is not all of $F(\mathbb{R})$. José above mentioned continuity, and so if you know what continuity is, you can use that. Another elementary reason, for instance, is the following. Take any nonzero $f\in P(\mathbb{R})$: then $f$ has only finitely many roots (in fact, if $f$ has degree $n$, then $f$ has at most $n$ roots, by the Fundamental Theorem of Algebra). So it suffices to choose a nonzero element $g\in F(\mathbb{R})$ that has infinitely many roots, i.e. whose graph crosses the $x$-axis infinitely many times. But I'm sure you can think of a nonzero function that crosses the $x$-axis infinitely many times: take e.g. $g(x) = \sin(x)$. This shows that $\sin$ is not in $P(\mathbb{R})$.