Basis for infinite-dimensional vector spaces Let $V$ be a vector space over a field $D$, and $U \subseteq V$ a subset.
Prove that the following are equivalent:


*

*For each $v \in V - \{0\}$ there exist unique $n \in \mathbb{N}$ and $u_1,...,u_n \in U$ pairwise disjoint and $k_1,...,k_n \in K  - \{0\}$ such that $v = \sum_{i=1}^{n}{k_i u_i}$

*$U$ is a maximal set of vectors linearly indipendent.
The fact that the second implies the first is obvious: my problem is regarding the opposite implication.
 A: If $\sum a_iu_i=0$ with $a_i\neq0$ then the representations in $1.$ would not be unique because $$v=\sum k_iu_i=\sum(k_i+a_i)u_i$$
Therefore $U$ are linearly independent.
If you add $v\neq0$ to the collection $u$, then $0=\sum k_iu_i-v$ is a non-trivial linear combination of $U\cup\{v\}$ that is equal to zero. Therefore you can't enlarge $U$ without making it linearly dependent.
A: You need to prove that $U$ is linearly independent, and that it is maximal (that is, it has no linearly independent proper superset).
For the first one, assume that $U$ is not linearly independent -- that is, there is a nontrivial linear relation. With a bit of shifting around you should be able to find a $v$ such that the combination of $n$ and $u_1,\ldots,u_n$ is not unique (as assumed).
For the second, assume that $W$ is a proper superset of $U$. Choose $v\in W\setminus U$, and apply the hypothesis about $U$ to this $v$. This will give you a nontrivial linear relation in $W$, so $W$ is not linearly independent.
