Linearly Indept set of n vectors implies size of vector space to be at least n. 
Theorem:
If a set of n vectors is linearly independent in a vector space, then the dimension of the space is at least n.

Proof:
Let $S = \{\vec{v}_{i}\}_{i=1}^{n}$ be a L.I set of n vectors in a vector space $V$.
Certainly, the S is a basis for the vector space $V$. The order of the set $S$ is $n$, and by definition, dim$\left(V\right) = |S| = n$.
It might suffice to show that the union of the set $S$ with another set $\bar{S}, \bar{S} \neq 0$
does not affect the span of $S$.
I.e, $S = S \cup \bar{S}$
Any help to take me further is greatly appreciated.
Thanks in advance.
 A: $S$ need not be a basis set. For example, in $\Bbb{R}^3$, we can have $S=\left\{\begin{bmatrix}1\\0\\0\end{bmatrix}, \begin{bmatrix}1\\1\\0\end{bmatrix}\right\}$ a set of $2$ linearly independent vectors but definitely it doesn't span $\Bbb{R}^3$. So it is not a basis of $\Bbb{R}^3$. However, one can have a basis of $\Bbb{R}^3$ which can contain (at least) these two vectors.  
In general, the best we can say is that one can have a basis $\mathcal{B}$ that can contain at least these $n$ linearly independent vectors (from $S$). Thus $\text{dim} \geq n$. 
A: Pick your set $S$ of linearly independent vectors in $V$, then if they don't span $V$, find a vector that is not in their span and add it to the set. You are still going to get a set of linearly independent vectors. Do this until you get a spanning set and then $S$ becomes a basis for $V$. In particular, you don't remove vectors from your initial set $S$ to get a spanning set. So the dimension of $V$ is at least $|S|$. Notice that it is not true that a linearly independent set of vectors is a basis for $V$. Take for example $V=\mathbb{R^3}$ and $S={(1,0,0),(0,1,0) \}$, $S$ is not a basis! 
