Subsets of linearly independent sets are linearly independent If $ S = \{ v_1,\dots,v_k\} \subset V (k \in \mathbb{N})$ is linearly independent and $T \subseteq S$, prove that $T$ is linearly independent.
 A: First of all, let's try to understand the definition of linear independence a bit better: we say the set $S$ is linearly independent if for any scalars $c_1,\dots,c_n$, the only way we can have
$$
c_1 v_1 + c_2 v_2 + \cdots + c_n v_n = 0
$$
is if $c_1 = c_2 = \cdots = c_n = 0$.
In order to understand the proof we're trying to go through, it might help to go through a bit of a "mini-proof".

Claim: If $T = \{v_1,v_2,v_3,v_4,v_5\}$ is linearly independent, then so is $S = \{v_1,v_2,v_3\}$.

Proof: We want to show that if $c_1,c_2,c_3$ are scalars such that $c_1 v_1 + c_2 v_2 + c_3 v_3 = 0$, then it must be the case that $c_1 = c_2 = c_3 = 0$. 
So, suppose that $c_1,c_2,c_3$ are such scalars.  We can then say that
$$
c_1 v_1 + c_2 v_2 + c_3 v_3 + (0)v_4 + (0)v_5 = 0
$$
However, the set $\{v_1,\dots,v_5\}$ is linearly independent.  So... [where do we go from here?]
A: A set of vectors $A$ is linearly independent if whenever $x_1,\ldots,x_n$ are in $A$ and $c_1,\ldots,c_n$ are scalars satisfying
$$
c_1x_1+\cdots+c_nx_n=0,
$$
then $c_1=\cdots=c_n=0$.
To show that $T$ is linearly independent, we fix vectors $x_1,\ldots,x_n$ in $T$ and $c_1,\ldots,c_n$ scalars satisfying
$$
c_1x_1+\cdots+c_nx_n = 0.
$$
Then in particular $x_1,\ldots,x_n$ are in $S$. Hence the fact that $S$ is linearly independent implies $c_1=\cdots=c_n=0$. This completes the proof.
