Prove that $\mathrm{span}(S) = S$ for a subspace $S$. Prove that if $S$ is a subspace of a vector space $V$, then $\mathrm{span}(S) = S$.
What I tried: I considered using the properties of vector spaces or maybe using an example where $S \subseteq \mathbb{R}^2$, but those strategies didn't amount to much. Any thoughts?
 A: $\mathrm{Span}(S)$ is the set of finite linear combinations of elements in $S$. In particular, every element in $S$ is in $\mathrm{Span}(S)$. Thus $S\subseteq\mathrm{Span}(S)$. Now, set $u\in\mathrm{Span}(S)$. Then $u=a_1s_1+...+a_ns_n$ for some scalars $a_i$ and $s_i\in S$. As $S$ is a subspace, it is closed under plus and product by scalar. Then $u\in S$. This proves $\mathrm{Span}(S)\subseteq S$.
A: Note that the span of a set $S\subset V$ is the intersection of all subspaces of $V$ that contain $S$, i.e. $$\operatorname{Span}(S) = \bigcap_{S\ \subset\ T\ \leqslant V}T. $$
The intersection of subspaces is again a subspace, so $$\bigcap_{S\ \subset\ T\ \leqslant V}T$$ is a subspace containing $S$, which implies that $$\operatorname{Span}(S)\subset \bigcap_{S\ \subset\ T\ \leqslant V}T. $$   But $\operatorname{Span}(S)$ itself is a subspace containing $S$, therefore
$$\operatorname{Span}(S)\supset \bigcap_{S\ \subset\ T\ \leqslant V}T. $$ 
It follows immediately that if $S\leqslant V$, then $\operatorname{Span}(S)=S$.
