Isomorphism between infinite dimensional vector spaces

Does defining an isomorphism $\theta: \mathbb R^{\mathbb N} \to$ {polynomials}  make sense? It does intuitively, but I am worried about the infinite nature. Thanks.

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The number of nonzero coefficients of a polynomial is by definition finite. The number of nonzero coordinates of a vector in $\mathbb R^\mathbb N$ can be infinite. Hence the "natural" morphism you might think about is not even defined. – Did Sep 11 '11 at 22:59
It depends on what you mean by $\mathbb{R}^{\mathbb{N}}$. The usual meaning is that it consists of all functions with domain $\mathbb{N}$ and image $\mathbb{R}$; this is essentially the vector space of real sequences. It is not, however, isomorphic to the space of polynomials, which consists only of those functions $\mathbb{N}\to\mathbb{R}$ with "finite support". So you won't be able to define an isomorphism. The dimension of the space of polynomials is $\aleph_0$, but the dimension of $\mathbb{R}^{\mathbb{N}}$ is $2^{\aleph_0}$. – Arturo Magidin Sep 11 '11 at 23:01

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 $\mathbb R[X]$.

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Thanks! Would you mind explaining why the dimensions are those given? – jon b Sep 11 '11 at 23:14
It may be confusing to refer to the ring of real polynomials, since we are dealing with vector spaces, and , AFAIK $\mathbb R^{\mathbb N}$ is not a ring. – gary Sep 11 '11 at 23:27
@gary: Sure, it is a ring with pointwise addition and multiplication. If $R$ is any ring, and $X$ is any nonempty set, then $R^X$ is a ring. However, you are correct that here we are thinking about vector spaces, so it is best not to use the term "ring". – Arturo Magidin Sep 11 '11 at 23:30
@jon b: The fact that the space of polynomials has dimension $\aleph_0$ can be verified by noting that the infinite countable set $\{1,x,x^2,\ldots,x^n,\ldots\}$ is a basis. The fact that the dimension of $\mathbb{R}^{\mathbb{N}}$ is $2^{\aleph_0}$ is more complicated; but basically, there are "too many" functions for the dimension to be denumerable. – Arturo Magidin Sep 11 '11 at 23:33
@Henning: A nice proof (which I first saw from Bill Dubuque) that the vector space $F^{\mathbb{N}}$ has dimension at least $|F|-1$ (which suffices here) is to consider for each $c\in F$, $c\neq 0$, the function $f_c(n) = c^n$; then show these are linearly independent. See for instance this – Arturo Magidin Sep 12 '11 at 0:21

• the set of sequences of real numbers of length $n$
The sequence $(1,1,1,1,\ldots)$ is one in which the sum of the squares is not finite, so it belongs to the fourth space but not to the third. The third one is an example of a Hilbert space. – Michael Hardy Sep 12 '11 at 14:51