Why can a vector from an infinite-dimensional vector space be written as finite linear combination? Suppose $V$ is a infinite-dimensional vector space over $\mathbb{R}$ or $\mathbb{C}$ and $\beta$ is a basis for $V$. I have seen the following claim:
For every $v$ $\in$ $V$, there exist $v_1,...,v_n \in \beta $ such that
$$v = \sum_{i=1}^{n} a_i v_i.$$
But when $v$ is a linear combination of infinite number of vectors in $\beta$, there does not exist such a integer $n$. How can we still choose vectors like this?
A related question is about $\textrm{span}(\beta)$. I know that $\textrm{span}$ is the set of all linear combinations of finite vectors in the set. But then it would not include vectors that are obtained by a linear combination of infinite number of vectors. How can this be the case?
 A: If all we know is that $V$ is a vector space, then "an infinite sum of vectors" is not necessarily defined. 
There are some contexts where infinite sums are defined.  For instance, if $V$ has a norm, then we have a notion of distances between vectors, and we can say that $\sum_{n=1}^\infty v_n = v$ if
$$
\lim_{N \to \infty} \left\| v - \sum_{n=1}^N v_n\right\| = 0.
$$
If $\{v_n\}$ is a "basis" in the sense that every $v$ can be written in the form $v = \sum_{n=1}^\infty a_nv_n$ (for some coefficients $a_n$), then we say that $\{v_n\}$ is a Schauder basis.  By contrast, the only kind of basis that makes sense without something like our additional norm structure (i.e. for an arbitrary vector space) is a Hamel basis.
A: For a general vector space $V$ it doesn't make sense to talk about infinite sums. I suppose you could define a norm on $V$, as it's a vector space over $\mathbb R$ or $\mathbb C$, but this doesn't generalize to other fields, such as $\mathbb Z / 2$. When it comes to general vector spaces, you can really only talk about finite sums. You can talk about infinite sums in, say, a Hilbert space, but that's a lot more structure.
The very definition of a basis $B \subseteq V$ is that every element in $V$ is a unique finite linear combination of elements in $V$, even if $B$ is infinite. For example, take $V = \{a \in \mathbb R^\mathbb N : a_n = 0 \text{ for all but finitely many } n\}$. Then letting $e_i(j)$ be 1 for $i = j$ and 0 otherwise, we have that $\{e_0, e_1, e_2, e_3, \dots\}$ is a basis for $V$. Although this is infinite, every element of $V$ is a finite linear combination of these basis elements. However, this set is not a basis for $\mathbb R^\mathbb N$. Indeed, the sequence $(1, 1, 1, 1, \dots)$ is not in the span. However, it is a theorem that all vector spaces have a basis, so there is a way to represent all of these sequences as a unique finite linear combination of other sequences. I can't write this basis down for you, as this theorem uses the axiom of choice (and is, in fact, equivalent to it). So if you accept the axiom of choice, your problem can be remedied by just saying "take some basis" without worrying about what it is. If you don't, then there will be some infinite dimensional spaces which don't admit a basis, so you can't always represent vectors in this way.
A: Infinite-dimensional vector space does not necessarily mean that it contains the "infinite" objects you think about. As others have pointed out, introducing infinity into the definitions would have to make you consider infinite series (where the arrangement of summands matters), limits etc. which are concepts that may work for real or complex numbers, but not for other objects that can nevertheless form vector spaces.
Thus when talking generally about vector spaces, we define the basis and span in terms of finite sums like the linear combination.
As an illustration, let's say that $β=\{e_i\}$. By definition, $span(β)=V$ cannot contain elements like $(1, 1, 1, ...)$, as it would require you to sum the whole (infinite) basis. You can have these elements in $V$, but they must be added to the basis if they cannot be formed using (finite) linear combination.
Speaking only in terms of finite sums is not only simpler, but also beneficial in this case, as it allows you to conceive more different infinite-dimensional vector spaces. You can have the vector space containing all polynomials ($β=\{1, x, x^2, x^3, ...\}$) but a polynomial of infinite degree doesn't exist, so when infinite sums are disallowed, the vector space contains only polynomials (and not something like $1 + x + x^2 + x^3 + ...$ of "infinite" degree). On the other end of the spectrum, there is the vector space of all sequences, but its basis must be uncountable (which is something that is perfectly valid for a change).
