Linearly independent subset of V ⊆ basis of V The Theorem states: 

Let V be a vector space that has a finite spanning set, and let S be a
  linearly independent subset of V. Then there exists a basis S' of V,
  with S ⊆ S'

I don't need a proof necessarily, just want to build an intuition for it. Also, how would I use this theorem to solve questions like this one:
a) Find a basis of R^4 containing the linearly independent set S = {(1,2,3,4),(-1,0,0,0)}.
 A: The idea is that $S$ is a basis for some subspace $W_1\subset V$, so if we choose any $v_1\in in V\backslash W_1$, then $S\cup\{v_1\}$ is linearly independent, and spans a subspace $W_2$ with $W_1\subset W_2\subseteq V$. If $W_2=V$ we are done, otherwise, we can find $v_2\in V\backslash W_2$ so that $S\cup\{v_1,v_2\}$ is linearly independent. Keep going until you have constructed a basis.
In your particular example, find a vector that is not in the span of $S$ (e.g. $(0,1,0,0)$ is an easy choice). Then $S'=\{(1,2,3,4),(-1,0,0,0),(0,1,0,0)\}$ is linearly independent and spans a $3$-dimensional subspace. Find a vector not in the span of $S'$ to form a set $S''$ that is a basis.
A: Take a basis $\{v_1,v_2,v_3,v_4\}$ of $\mathbb{R}^4$ (the standard basis, for instance). Consider the set $S_1=S\cup\{v_1\}$. Is it linearly independent? If it is, keept it; otherwise, call $S_1$ to the set $S\cup\{v_2\}$ and so on. At a certain point, you'll have a linearly independent set $S_1$ with $3$ elements. Now, you start all over again (except that there's no need to try again an element of $\{v_1,v_2,v_3,v_4\}$ which has already been tested). So, suppose that $S_1=S\cup\{v_2\}$ (and this will be true if $\{v_1,v_2,v_3,v_4\}$ is the standard basis). Now, you start all over again, doing $S_2=S_1\cup\{v_3\}$. When you get a set with $4$ linearly independent vectors, that will be your basis.
