# Does the linear independence of given countable set can be deduced by that of its finite subsets?

In p.43 example 16 of Hoffman/Kunze's Linear Algebra(2nd Edition), there's the following argument:

Let $V$ be the vector space of polynomial functions over the complex numbers and let $f_k(x)=x^k$ and we want to show $B=\{f_0, f_1, f_2, \cdots \}$ is the basis for $V$. Clearly $B$ spans $V$, so it remains to show $B$ is linearly independent.

And here comes the part where I got confused:

To show this, it is enough to show that each finite subset of $B$ is linearly independent.

What I want to ask you is :

1. Does this hold by the fact that the set whose cardinality is strictly less than $\aleph_0$ is only finite set?
2. If so, what if the vector space $V$ had uncountable dimension? To show the subset $S$ of $V$ is linearly independent, is it enough to show the proper subsets of $S$ are linearly indepent?

A subset $S \subseteq V$ is linearly independent by definition if given any finite subset $F \subseteq S$, the only solution of the equation $\sum_{v \in F} a_v \cdot v = 0$ is $a_v = 0$ for all $v \in F$. In particular, this implies that a subset $S \subseteq V$ is linearly independent if and only if any finite subset of $S$ is linearly independent, irregardless of the cardinality of the dimension of $V$.
by definition: a vectors set $A$ of $V$ is a free system if and only if each finite subset $S$ of $A$ is free ( Independent of the nature of A countable or not).
when $A$ is countable, to show that $A$ is free, it is specified in sub set $S$, whose cardinal $n$, that way, when $n$ runs $N$, all finite subsets of $A$ will be taken.