If a set spans the vector space, and is contained in the linear span of another set, then the second set also spans 
Let $S_1$ and $S_2$ be subsets of a vector space $V$. Assume that $\text{Span} (S_1) = V$ and that every vector in $S_1$ is in $ \text{Span} (S_2)$. Show that $V = \text{Span}(S_2)$

So I said let $x$ be a vector in $\text{Span}(S_2)$
This implies that $x$ is a linear combination of vectors
Which would then have to mean $x$ is in $\text{Span} (S_1) $
And since we know $\text{Span} (S_1) = V$ then $\text{Span} (S_2) = V$
I know the notation could be better, but I am on a tablet and my latex is bad
 A: Since $S_1\subseteq\text{span}(S_2)$, $V=\text{span}(S_1)\subseteq\text{span}(S_2)$; so $V=\text{span}(S_2)$ since $\text{span}(S_2)\subseteq V$.
A: Let me give a different answer.

Let $S_1$ and $S_2$ be subsets of a vector space $V$. Assume that $\text{Span} (S_1) = V$ and that every vector in $S_1$ is in $ \text{Span} (S_2)$. Show that $V = \text{Span}(S_2)$

I am assuming everything finite here, i.e., $S_1 = \{v^{(1)}_1,\ldots, v^{(1)}_n\}$ and $S_2 = \{v^{(2)}_1,\ldots, v^{(2)}_m\}$.
Proof:
Let $w$ in $V$.
Since $\text{Span} (S_1) = V$,
$$w = \sum_{i=1}^n a_i v^{(1)}_i$$
But $v^{(1)}_i \in \text{Span}(S_2)$, so
$$v^{(1)}_i = \sum_{j=1}^m b_{ij} v^{(2)}_j$$
Using both we get
$$w = \sum_{i=1}^n a_i \sum_{j=1}^m b_{ij} v^{(2)}_j \\
= \sum_{j=1}^m \sum_{i=1}^n a_i b_{ij} v^{(2)}_j$$
But now $w$ can be written as a linear combination of $v^{(2)}_j \in S_2$:
$$w = \sum_{j=1}^m c_i v^{(2)}_j$$
In which $c_i = \sum_{i=1}^n a_i b_{ij}$.
Since $w$ was arbitrary, $V = \text{Span}(S_2)$.
$\square$
