The definition of span

Remember that a subspace by definition is closed with respect to vector addition. That means that every subspace which contains $S$ necessarily contains every linear combination of elements of $S$. In turn then, the intersection of all such subspaces is exactly the set of all linear combinations of vectors in $S$.

Let $${\alpha _1},...,{\alpha _n} \in V\;$$ then we define the space spanned by $${\alpha _1},...,{\alpha _n}$$ as $$S\left( {{\alpha _1},...,{\alpha _n}} \right) = \left\{ {\sum\limits_{i = 1}^n {{c_i}{\alpha _i}\;|\;{c_i} \in \mathbb{R}} } \right\}$$ which is the set of all linear combinations of the vectors in this subspace.

This Set is a subspace of V $$S\left( {{\alpha _1},...,{\alpha _n}} \right) \subset V$$

The intersection property says that the intersection of subspaces in V is itself a subspace of V

So then we can define S as an intersection of an arbitrary family of subspaces of V which contain S $$S\left( {{\alpha _1},...,{\alpha _n}} \right) = \mathop \cap \limits_{m \in F} {W_m}$$

Forward: Let us consider the first definition of span. We will prove that it implies the second definition of span.

Theorem 1: The span of any subset $S$ of a vector space $V$ is a subspace of $V$. Moreover, any subspace of $V$ that contains $S$ must also contain span of $S$. (Referred from Friedberg)

Proof: This result is immediate if $S=\phi$, because $span(\phi)=\{0\}$, which is a subspace that is contained in any subspace of $V$.

If $S\neq \phi$, then $S$ contains a vector $z$. So, $0z=0$. Therefore, $0 \in span(S)$. Let $x,y \in span(S)$. Then $\exists$ vectors $u_1,u_2,...u_m,v_1,v_2,...v_n$ and scalars $a_1,a_2,...a_m,b_1,b_2,...b_n$ such that $x=a_1u_1+a_2u_2+...+a_mu_m$ and $y=b_1v_1+b_2v_2+...+b_nv_n$

Then, $x+y=a_1u_1+a_2u_2+...+a_mu_m+b_1v_1+b_2v_2+...+b_nv_n$ and for any scalar $c$, $cx=(ca_1)u_1+(ca_2)u_2+...+(ca_m)u_m$; which are clearly linear combinations of the vectors in $S$. Thus, $x+y,cx \in span(S)$ i.e. $span(S)$ is a subspace of $V$.

Now let $W$ denote any subspace $V$ that contains $S$. If $w\in span(S)$, then $w=c_1w_1+c_2w_2+...+c_kw_k$ for some vectors $w_1,w_2,...w_k \in S$ and some scalars $c_1,c_2,...c_k$. Since $S \subseteq W$, we have $w_1,w_2,...w_k \in W$. Since $W$ is a subspace of $V$, addition and scalar multiplication is closed in $W$. Therefore, $w=c_1w_1+c_2w_2+...+c_kw_k \in W$. Since $w$ is any arbitrary vector in $span (S)$ that belongs to $W$, therefore $span(S) \subseteq W$.Q.E.D.

Since any subspace of $V$ that contains $S$ must also contain span of $S$, therefore the intersection $W$ of all subspaces of $V$ which contain $S$ gives us $span(S)$.Q.E.D.

Converse: Now we will consider the second definition of span and prove that it implies the first definition of span.

Let $S = \{w_1,w_2,...,w_k\}$ be a set of vectors in a vector space $V$. Then $span(S)=\cap_i W_i$, such that $S \subseteq W_i$ and $W_i$ is subspace of $V$ for all $i$.

As intersection of a collection of subspace is also a subspace, therefore $span(S)$ is a subspace of $V$.

Since $S \subseteq W_i$ for all $i$ and $w_1,w_2,...w_k \in S$, therefore $w_1,w_2,...w_k \in span(S)$

Since $span(S)$ is a subspace of $V$, therefore it is closed under addition and scalar multiplication. Thus, $w=c_1w_1+c_2w_2+...+c_kw_k \in span(S)$ for scalars $c_1,c_2,...c_k$. Since $w$ is any arbitrary vector in $span (S)$, therefore $span(S)$ is the set of all linear combinations of vectors in $S$.Q.E.D.

Hence, both the definitions are equivalent.