TITLE: Definition of basis in infinite-dimensional vector space
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: 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 $x\in S$”.