Describe the Bidual of Real Polynomials Trying to understand (bi)duals of infinite-dimension vector spaces, I stumbled over the very concrete example of $\mathbb{R}[X]$, the (formal) polynomials over $\mathbb{R}$ or, equivalently, the space of $\mathbb{R}$-sequences with almost all values zero, interpreted as a vector space. Its dual vector space is $\mathbb{R}[X]^{*} = [\mathbb{R}[X] \to^\text{lin} \mathbb{R}]$ which—according to Wikipedia—is isomorphic to $\mathbb{R}[\![X]\!]$, the vector space of formal power series over $\mathbb{R}$ or, equivalently, the space of $\mathbb{R}$-sequences.
I succeeded to find an explicit isomorphism (and its inverse) myself and prove that they are in fact linear and inverses.
Where I really struggle is finding a concrete representation of
$$
  \mathbb{R}[\![X]\!]^{*}
= [\mathbb{R}[\![X]\!] \to^\text{lin} \mathbb{R}]
\simeq [[\mathbb{R}[X] \to^\text{lin} \mathbb{R}] \to^\text{lin} \mathbb{R}],
$$
 i.e. the dual of $\mathbb{R}[\![X]\!]$ i.e. the bidual of $\mathbb{R}[X]$. Does it even exist or is it rather a happy coincidence that the dual of $\mathbb{R}[X]$ has such a nice representation? I know that my question is very specific and vague.
 A: I'm no algebraist, but hopefully I can provide a helpful comment. What makes the dual of $R[X]$ nice is that $R[X]$ has a nice countably infinite Hamel basis $B = \{X^0,X^1,X^2,X^3,\ldots\}$: every element of $R[X]$ can be written uniquely as a (finite) linear combination of elements in $B$. An element of the dual of $R[X]$ can be described as a formal infinite linear combination of dual elements of $R[X]$. That is, if we define a "dual basis" $B^* = \{ Y^0, Y^1, Y^2,\ldots \}$ where $Y^i(X^j) = \delta_{ij} = 1$ if $i = j$ and $0$ otherwise, then the set of formal linear combinations $R[X]^* = \{ \sum_{j\in\mathbb{N}_0} c_jY^j : c_j \in {\mathbb R} \}$ is the dual space of $R[X]$. The fact that the formal linear combination is infinite is mollified by the fact that any element of $R[X]$ is a traditional finite linear combination of elements in $B$, so only finitely many of the $Y^i(X^j)$ become nonzero in the final summation.
We can proceed in the same way for any vector space $V$. Given a Hamel basis $B$ for $V$, we construct a "dual basis" $B^*$ and we can identify $V^*$ with infinite linear combinations of elements in $B^*$.
In general, we can use the axiom of choice to show any vector space $V$ has a Hamel basis, but in general we need the axiom of choice to construct these bases---there exist vector spaces for which its impossible to write down a Hamel basis using only countably many operations. This is where my ignorance of algebra starts to really hamper me, but the space $R[\![X]\!]$ may not have a Hamel basis we can construct without the axiom of choice, in which case this percludes using the formal infinite linear combination construction you use to construct an explicit representation of $R[X]^*$. Every time you take the (algebraic) dual of an infinite dimensional vector space, the dimension of the dual is a strictly larger infinite cardinal number, so duals and bi-duals and tri-duals become increasingly ugly.
