TITLE: Span of a subset of a vector space is the smallest subspace containing that set
QUESTION [15 upvotes]: To Prove:
If  $S=[{v_1,v_2,...,v_k}]$ is a subset of vector space $V$. Then $span(S)$ is the smallest subspace of $V$ containing set $S$.
I know that $L[S]$ is a subspace of $V$. 
But in most arguments for proving the $L[S]$ is the smallest subspace containing $S$ , I find that if $W$ is another subspace of $V$ containing $S$ then, proving $S \subset W$ means $S$ is the smallest. I couldn't understand that if $S \subset W$ proves that $L[S]$ is the smallest containing $S$. Please elaborate.

REPLY [12 votes]: Theorem: Let $S$ be a subset of a vector space $V$. Then $\text{span}(S)$ is the smallest vector subspace of $V$ which contains $S$.

What kind of set $T$ would disprove the theorem? It would have to be (1) a vector subspace which (2) contains every vector in $S$, and which (3) is smaller than span(S) so that $T\subsetneq\text{span}(S)$.
Let's suppose $T$ has the first two qualifications. We'll show that it can't have the third qualification; that's enough to establish the proof. In particular, suppose: 

$T$ is not just a set; it's a vector subspace. This means that $T$ is closed under linear combinations. (every linear combination of vectors in $T$ also belongs to $T$.)
$T$ contains every member of $S$. ($T \supseteq S$).

By the first two properties, $T$ contains every vector in $S$ and $T$ is closed under linear combinations, so it follows that $T$ contains every linear combination of vectors in $S$.  But $\text{span}(S)$ is by definition the set of all linear combinations of vectors in $S$.  Therefore $T\supseteq \text{span}(S)$.  Therefore $T$ is at least as large as $\text{span}(S)$, so the theorem is true.