# If $V$ is a vector space, that what is $V^\mathbb{N}$?

Notation question (I believe the notes that I'm reading uses pretty common notation):

Let $V$ be a vector space over a field $K$. What is $V^\mathbb{N}$? Is it a vector space of infinite dimensions? What are the elements of $V^\mathbb{N}$? Are they infinite length tuples?

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BTW I'd have preferred to ask this question on chat.math.stackexchange.com but it's not setup yet. –  user2468 Oct 26 '10 at 15:47

In my opinion, this notation is ambiguous (edit: but maybe I'm wrong). It could mean two things: it could mean the space of functions $\mathbb{N} \to V$, which is a vector space in the obvious way. This is the countable product in the category of vector spaces. This vector space is very large; in particular, it has uncountable dimension and it is not spanned by the functions $e_i$ which are equal to $1$ at $i$ and equal to $0$ elsewhere.

Or it could mean the space of functions $\mathbb{N} \to V$ which are zero except at finitely many places, which is also a vector space in the obvious way, but countable-dimensional, and this is the vector space spanned by the $e_i$. This is the countable coproduct in the category of vector spaces. (Note that it is not necessary to distinguish between finite products and finite coproducts because they are the same.)

If I had to guess, I would guess that it means the former. Can you give a link to the notes? It might be possible to figure out which one is meant from context.

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Unfortunately the notes are under protected access within my campus :[ (edit:oops return) Thanks. I check both function spaces and infer which is consistent with the notes. –  user2468 Oct 26 '10 at 15:42
I would guess the former too. Most of the time I see the notation $V^\mathbb{N}$ it means "the space of sequences taking values in $V$". –  Willie Wong Oct 26 '10 at 15:44
@Qiaochu: I don't think $V^{\mathbb{N}}$ is used for the coproduct, though I have seen $V^{(\mathbb{N})}$ for the elements of $V^{\mathbb{N}}$ of finite support... –  Arturo Magidin Oct 26 '10 at 15:59
@Qiaochu: Could you elaborate on why the former has uncountable dimension? –  Rahul Oct 26 '10 at 18:25
@Arturo: Thanks for the link. I'm not very well-versed in infinite-dimensional vector spaces, so it will take me a while to digest that proof. @Qiaochu: Ah, I see; I was forgetting that the span is composed of linear combinations, which by definition involve only finitely many vectors. –  Rahul Oct 26 '10 at 23:04

It would be the vector space consisting of an infinite product V × V × ... of copies of V.

More generally: for two sets A and B, AB often denotes the set of functions f : B → A. (The only exception I'm aware of is when the sets A and B happen to be ordinals, in set theory; then a different notion of exponentiation is often used.)

How are these two concepts related? Well: as a vector space, V is also a set of vectors. And so a function f : ℕ → V is essentially a block-vector description of a vector v = (f(0), f(1), ...) ∈ V × V × ... .

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surely in the beginning of the second paragraph you meant $f:B\to A$. –  Willie Wong Oct 26 '10 at 15:48
@Willie Wong: that's right, silly mistake. –  Niel de Beaudrap Oct 26 '10 at 15:50

I'd think of it as the set of sequences of elements of $V$, i.e.

$$V^\mathbb{N} = \prod_{n=1}^\infty V.$$

But I don't think there is a really nice description. In the case where $V$ is finite-dimensional, it'd just be $K^\mathbb{N}$ in a clever (and non-canonically isomorphic) disguise, but if $V$ were to be infinite-dimensional, $V^\mathbb{N}$ could be rather ugly. E.g. $\ell_1^\mathbb{N}$: sequences of absolutely summable sequences. Then a sequence of sequences would converge to a sequence iff it converged coordinatewise. Complicated and very confusing.

Following Munkres' "Topology", I prefer to use $V^\infty$ for set of sequences in $V$ for which only finitely many elements are nonzero, i.e.

$$V^\infty = \bigoplus_{n=1}^\infty V = \left\lbrace (v_n) \in V^\mathbb{N} \;{\large\mid}\; \text{v_n = 0 from some index onwards}\right\rbrace.$$

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