Proof verification: polynomials $\mathbb R[X]$ are a vector space that is not isomorphic to its dual

Question

I have never seen an elementary proof of the fact that $V$ need not be isomorphic to $V^*$ that does not require some set theoretic background. I came up with this [most likely incorrect] argument, which does not seem to depend so much on set theoretic arguments, except for basic ideas of cardinality. I wish to have this particular proof vetted.

Consider the vector space of polynomials $V \equiv \mathbb R[X]$ as an $\mathbb R$ vector space. The set $B_V \equiv \{x^i : i \in \mathbb N \}$ is a basis for the vector space $V$. Given any polynomial $p(x) \in \mathbb R[X]$, since the polynomial $p$ only has finitely many non-zero coefficients. Thus $p(x)$ must be of the form $p(x) = \sum_{i \in \text{nonzero-powers}(p)} a_i x^i$ where the index set $\text{nonzero-powers}(p)$ has finite cardinality. Hence we can write any polynomial $p(x)$ as a finite linear combination of elements from the set $B_V$.

Next, consider the dual space $V^* \equiv \{ f : \mathbb R[X] \rightarrow \mathbb R \mid f \text{ is a linear function} \}$. We have the elements $eval_r$ which evaluate a polynomial at point $r \in \mathbb R$ as elements of $V^*$. More formally, $eval_r(p) \equiv p(r); \forall r \in \mathbb R, eval_r \in V^*$.

All the $eval_r$ are linearly independent. Intuitively, this is because we cannot pin down the value of all polynomials by evaluating them at some finite number of points.

More formally, suppose that we have that $\sum_{i \in I} a_i eval_{r_i} = 0$ for some finite index set $I$. So this gives us a way to extrapolate $eval_{i_0}$ from the other $eval_i$. However, this is absurd, since the value of a polynomial of degree $2|I|$ is not determined by its value at $|I|$ points. Hence all the $eval_r$ are linearly independent.

This means that we have a linearly independent set $L_{V^*} \equiv \{ eval_r : r \in \mathbb R \}$ whose cardinality is that of $|\mathbb R|$.

Wrapping up, we have that the basis of $V$, $B_V$ has cardinality $|\mathbb N|$. A linearly independent set of $V^*$, whose cardinality is a lower bound on the cardinality of $V^*$, has cardinality $|\mathbb R|$. Hence the vector spaces cannot be isomorphic since the cardinality of their bases are different.

Is this correct?

Answer 1

This is indeed correct. Note that you don't need to talk about a basis of $V^*$ (something whose existence depends on the Axiom of Choice): simply the fact that $V^*$ has an uncountable linearly independent set, while $V$ does not, establishes that they can't be isomorphic.

Answer 2

Some comments. The dual space $V^{\ast}$ can be identified with real formal power series $\mathbb{R}[[Y]]$; the identification sends a formal power series $\sum a_i Y^i$ to the linear functional

$$\mathbb{R}[X] \ni \sum b_i X^i \mapsto \sum b_i a_i \in \mathbb{R}.$$

With respect to this identification the evaluation homomorphisms correspond to the formal power series $\frac{1}{1 - rY}$. There are many ways to show that these formal power series are linearly independent; maybe the shortest one is to notice that, interpreting them as genuine functions of a real-valued variable $Y$, any nontrivial linear combination of them has a pole somewhere (and hence isn't identically zero).

Any other method of writing down an uncountable family of linearly independent formal power series also constitutes a proof that $V^{\ast}$ is not isomorphic to $V$; another very similar family is the family $\exp (r Y)$. A cute way to show that these power series are linearly independent is to repeatedly differentiate any nontrivial linear combination of them; you can also observe that, again interpreting them as genuine functions of a real-valued variable $Y$, in any nontrivial linear combination one term has the uniquely largest growth rate as $Y \to \infty$ and hence eventually dominates the others.