Definition of basis in infinite-dimensional vector space I am struggling to understand the definition of a basis in an infinite dimensional vector space. Specifically, the definition I know says: A subset $B$ of a vector space $V$ is a basis for $V$ if every element of $V$ can be written in a unique way as a finite linear combination of elements from $B$.
However, for any non-empty subset $X$ of a vector space $V$, the zero element of the space can be written in more than one way as a finite linear combination of elements from $X$. For example, $0 = 0v = 0w$, where $v \neq w$ are from $X$. So therefore, no subset $X$ of a vector space $V$ could be a basis for $V$.
What am I missing? What exactly does the definition mean?
 A: $\newcommand{\Reals}{\mathbf{R}}$Let $(V, +, \cdot)$ be a real vector space. One commonly says an ordered set $S = (v_{i})_{i \in I}$ indexed by a set $I$ is a basis of $V$ if the following hold:


*

*$S$ spans $V$, i.e., for every $v$ in $V$, there exists a function $c:I \to \Reals$, non-zero for only finitely many $i$, such that $v = \sum_{i} c(i) v_{i}$.

*$S$ is linearly independent, i.e., if $c:I \to \Reals$ is a function, non-zero for only finitely many $i$, and if $0 = \sum_{i} c(i) v_{i}$, then $c(i) = 0$ for all $i$.
In this framework, "uniqueness of representation" is a (simple) theorem: If $v \in V$, there exists a unique function $c:I \to \Reals$, non-zero for at most finitely many $i$, such that $v = \sum_{i} c(i) v_{i}$.
The point is, you can (and for uniqueness, should) view a linear combination from $S$ as a sum over all elements of $S$, but in such a way that at most finitely summands have a non-zero coefficient.
A: The language used is a bit sloppy, but it's common not to be too fussy in these definitions.
Let $S$ be a subset (finite or infinite, it doesn't matter) of $V$. A choice of coefficients for $S$ is a function $f\colon S\to F$ (where $F$ is the base field) such that $\sigma(f)=\{x\in S:f(x)\ne0\}$ is finite.
We observe that, if $T$ is a finite subsets of $S$ such that $\sigma(f)\subseteq T$, then
$$
\sum_{x\in\sigma(f)}f(x)x=\sum_{x\in T}f(x)x
$$
with the convention that
$$
\sum_{x\in\emptyset}f(x)x=0
$$
Since the summation doesn't depend on the finite subset we choose, we set, for a choice of coefficients $f$,
$$
\sum_{x\in S}f(x)x=\sum_{x\in\sigma(f)}f(x)x
$$
and call this vector a linear combination of $S$. Note that the summation is actually finite and we can use whatever subset $T$ we want, instead of $\sigma(f)$, provided $\sigma(f)\subseteq T$ and $T$ is finite. This is mostly useful for doing computations with choices of coefficients, when other properties are being investigated.
Then we call $S$ a basis for $V$ if


*

*For every $v\in V$, there exists a choice of coefficients $f$ for $S$ such that
$$
\sum_{x\in S}f(x)x=v
$$
(we can abbreviate this condition by saying that $S$ is a spanning set for $V$).

*For every $v$, if $f$ and $g$ are choices of coefficients for $S$ and
$$
\sum_{x\in S}f(x)x=\sum_{x\in S}g(x)x
$$
then $f=g$ (we can abbreviate this condition by saying that $S$ is linearly independent).
Note that condition 2 can be rewritten as “If $f$ is a choice of coefficients for $S$ and
$$
\sum_{x\in S}f(x)x=0
$$
then $f(x)=0$, for every $x\in S$”.
A: The expression for $0$ you written are not two; they are same:
$$0=0v=0v+0w \mbox{ and } 0=0w=0v+0w .$$

Let $B$ be a subset of $V$. Let $v\in V$ be written as a finite linear combination of elements of $B$:
$$v=a_1v_1+a_2v_2+\cdots + a_nv_n =b_1w_1+b_2w_2+\cdots + b_mw_m, \,\,\,\, 
v_i,w_j\in B.$$
Question: When do we say that these expressions are same? 
To answer this, note that $v$ is expressed as linear combination of elements of the set $\{v_1,\cdots,v_n\} \cup \{w_1,\cdots,w_m\}$, and some $v_i$'s and $w_j$'s could be same. We can rewrite the two expressions as below:
$$ v = a_1v_1+\cdots + a_nv_n + 0w_1 + \cdots + 0w_m.$$
and 
$$ v = 0v_1+\cdots + 0v_n + b_1w_1 + \cdots + b_mw_m.$$
Since some $v_i$'s and $w_j$'s could be same, so let 
$$\{v_1,\cdots, v_n\} \cup \{ w_1,\cdots , w_m\}=\{u_1,u_2,\cdots, u_l\}$$
where all the $u_i$'s are distinct. Then the above two expressions for $v$ can be expressed in terms of $u_i$'s only, using rules in scalar multiplication:
$$v=c_1u_1 +\cdots + c_lu_l = c_1'u_1 + \cdots + c_l'u_l.$$
Now these two expressions for $v$ are said to be same if $c_i=c_i'$ for all $i$.
Illustration: Let $v=a_1v_1+a_1v_2$ and $v=b_1w_1 + b_2w_2$, and suppose, $v_2=w_2$  Then the set $\{v_1,v_2\} \cup \{w_1,w_2\}$ contains only three elements: $v_1,v_2,w_1$; relabel them as 
$$v_1=u_1, \,\,\,\, v_2=u_2, \,\,\,\, w_1=u_3.$$
Note that $w_2=v_2$ and $v_2$ is labeled as $u_2$, so $w_2=u_2$. 
The two expressions of $v$ now become:
$$v=a_1v_1 + a_2v_2 =a_1u_1 + a_2u_2= a_1u_1 + a_2u_2 + 0u_3$$
and 
$$ v =b_1w_1 + b_2w_2= b_1u_3 + b_2u_2  = 0u_1 + b_2u_2 + b_1u_3.$$
Now can you say when  the last two expressions for $v$ are same?
A: When you say "in a unique way", it implies that you disregard the zeroes. Otherwise of course no space would have a basis, infinite-dimensional or not (except the zero space...).
This being said, if you want a more subtle definition : a set $S\subset X$ is a basis of $X$ if for any vector space $Y$ and any function $f:S\to Y$ there is exactly one linear map $X\to Y$ that extends $f$.
