Finitely Spanned Vector Space Ok my textbook says that $\mathbb{F^{Z^+}}$ is not finitely spanned. The argument is obvious, there are no finite list whose span gives $\mathbb{F^{Z^+}}$. But when we talk about vector spaces we refer to two sets. The sets of vectors $V$ and a field $F$. Why can't we take a finite list, any list $\{1,2,3\}$ Any positive integer is a linear combination of this sublist. So what's the problem in my understanding?
 A: By $ \mathbb{F}^{\mathbb{Z}_{+}} $, I believe you mean the set of all functions from $ \mathbb{Z}_{+} $ into $ \mathbb{F} $, which is simply the set of all sequences in $ \mathbb{F} $.
This is an infinite-dimensional vector space. To see this, simply note that the following is an infinite linearly independent subset of $ \mathbb{F}^{\mathbb{Z}_{+}} $:
$$
S := \{ (1_{\mathbb{F}},0,0,0,\ldots),(0,1_{\mathbb{F}},0,0,\ldots),(0,0,1_{\mathbb{F}},0,\ldots),\ldots \}.
$$
Using Zorn’s Lemma to extend $ S $ to a basis of $ \mathbb{F}^{\mathbb{Z}_{+}} $, we see that $ \mathbb{F}^{\mathbb{Z}_{+}} $ must be infinite-dimensional. As the dimension of a vector space is an invariant, once you have an infinite basis with cardinality $ \kappa $, all other bases must also be infinite with cardinality $ \kappa $.
Note: The linearly independent set $ S $ that I have provided above is not a basis of $ \mathbb{F}^{\mathbb{Z}_{+}} $ because any finite linear combination of elements of $ S $ only gives you a finite sequence (i.e., a sequence that consists only of $ 0 $’s after some point). Therefore, an actual basis of $ \mathbb{F}^{\mathbb{Z}_{+}} $ is more complicated.
