Span of a subset of a vector space is the smallest subspace containing that set 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. 
 A: Here is your statement written slightly different. Maybe that will help. 
Theorem: If $W\subset V$ is a subspace, such that $v_1,…,v_k∈W,$ then $\text{span}(v_1, …,v_k)\subset W$.
Proof: Since $v_1,…,v_k∈W$ and $W$ is a subspace all linear combinations $$α_1v_1 + … + α_kv_k∈W$$
Since $\text{span}(v_1,…,v_k)$ contains (only) these linear combination it follows $\text{span}(v_1,…,v_k)\subset W$. 

Written in words that theorem states, that any subspace $W$, that exists and contains $v_1,…,v_n$, also contains $\text{span}(v_1,…,v_k)$. 
Hence $W$ is larger (or equal) to $\text{span}(v_1,…,v_k)$, since it contains $\text{span}(v_1,…,v_k)$ and could contain some more elements.  
And since any other subspace $W$ is larger (or equal) it follows that $\text{span}(v_1,…,v_k)$ the smallest subspace. 
That is like saying: Any number in $ℕ∪\{0\}$ is larger (or equal) to $0$, hence $0$ is the smallest number in $ℕ∪\{0\}$.
Or even closer to the original problem: Any subset $M⊂(ℕ∪\{0\})$ with $0∈M$ is larger or equal to $\{0\}$, hence $\{0\}$ has to be the smallest subset of $ℕ∪\{0\}$ that contains $0$. 
A: 
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.
