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.
Isomorphism between infinite dimensional vector spaces
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4It 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 $2^{\aleph_0}$. – 2011-09-11
2 Answers
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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0@Arturo, yes, that is nicer than what I could cobble together myself at short notice :-) – 2011-09-12
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.
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0The 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. – 2011-09-12