Isomorphism between infinite dimensional vector spaces Does defining an isomorphism $\theta: \mathbb R^{\mathbb N} \to \{\text{polynomials}\}$ make sense? It does intuitively, but I am worried about the infinite nature. Thanks.
 A: Think about these:


*

*the set of sequences of real numbers of length $n$

*the set of finitely long sequences of real numbers

*the set of infinite sequences of real numbers such the sum of their squares is finite

*the set of all infinitely long sequences of real numbers


The first is a finite-dimensional vector space.  The next three are infinite-dimensional vector spaces.  They're not all the same space.  When you understand the difference between them, you're on your way to understanding infinite-dimensional vector spaces.
A: If by $\mathbb R^{\mathbb N}$ you mean the set of all infinite sequences of real numbers (which is the standard meaning for that notation), then there isn't any isomorphism onto $\mathbb R[X]$ (the set of real polynomials in one variable). The two vector spaces have different dimension -- $\mathbb R^{\mathbb N}$ is $2^{\aleph_0}$-dimensional, whereas $\mathbb R[X]$ is only $\aleph_0$-dimensional.
($\mathbb R[X]$ has dimension $\aleph_0$ because the countably many polynomials $1$, $X$, $X^2$, $X^3$, ... form a basis. $\mathbb R^{\mathbb N}$ has dimension at most $2^{\aleph_0}$, because that's how many elements the vector space has. I don't have a slick argument that its dimension is at least $2^{\aleph_0}$ (but see the comments where Arturo gives one), but somewhat indirectly: $\mathbb Q^{\mathbb N}$ must have dimension $2^{\aleph_0}$ over $\mathbb Q$, because a basis smaller than that wouldn't be able to produce enough elements by finite linear combinations. So take a basis for $\mathbb Q^{\mathbb N}$ and look at the corresponding elements of $\mathbb R^{\mathbb N}$. The resulting set will still be linearly independent in $\mathbb R^{\mathbb N}$ -- any nontrivial linear relation in $\mathbb R^{\mathbb N}$ would create at least one nontrivial relation in $\mathbb Q^{\mathbb N}$, when the coefficients are expanded in coordinates under a basis for $\mathbb R$ as a vector space over $\mathbb Q$. Therefore $\mathbb R^{\mathbb N}$ has dimension at least $2^{\aleph_0}$ over $\mathbb R$).
The (proper) subspace of sequences where there are only finitely many nonzero elements -- sometimes notated $\mathbb R^\infty$ -- is naturally isomorphic to $\mathbb R[X]$.
