If $S$ is a subset of a vector space $V$, then $\text{span}(S)$ equals the intersection of all subspaces of $V$ that contain $S$. My introductory linear algebra textbook claims the following:

If $S$ is a subset of a vector space $V$, then $\text{span}(S)$ equals the intersection of all subspaces of $V$ that contain $S$.

I understand the aforementioned individual concepts, such as subsets, vector spaces, subspaces, and span, but I do not understand what is meant by, "$\text{span}(S)$ equals the intersection of all subspaces of $V$ that contain $S$." In addition, it seems to me that this statement is false: $\text{span}(S)$ does not necessarily have to equal the intersection of all subspaces of $V$ that contain $S$.
I would greatly appreciate it if someone would please take the time to elaborate on this concept and clarify what the textbook is saying. Please refrain from introducing more complex concepts from linear algebra in any explanation.
 A: $\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$There is a recurring situation in algebra, when you try and define 

the subobject $\Span{S}$ of an object $X$ generated (spanned) by a subset $S \subseteq X$,

where $\text{object} \in \{ \text{vector space}, \text{group}, \text{ring}, \dots\}$.
$\Span{S}$ is usually defined as 

the smallest (with respect to inclusion) subobject of $X$ which contains $S$.

The trouble is, just saying a magic formula like this does not guarantee that this objects exists.
So you usually try and do two things.
$$\tag{existence}\text{Prove that this subobject exists.}$$
$$\tag{description}\text{Describe it in terms of the elements of $S$.}$$
A convenient way of addressing (existence), which has usually no impact on the much more useful (description), though, is to prove the following.


*

*The intersection of any collection of subobjects is a subobject.

*In particular, the intersection $Y$ of all subobjects of $X$ containing $S$ is a subobject of $X$ containing $S$.

*If $Z$ is a subobject of $X$ containing $S$, then $Z$ is one of the terms of this intersection, and thus contains $Y$.

*Therefore $Y$ is the smallest (with respect to inclusion) subobject of $X$ containing $S$.


In the case of vector spaces, of course the useful bit (description) is that $\Span{S}$ is the set of (finite) linear combinations of elements of $S$.
A: The vector space $\operatorname{span}(S)$ contains $S$ and consists only of linear combinations of elements of $S$. Closure under addition and scalar multiplication means that for any vector space $W\subseteq V$, if $S\subseteq W$ then also $\operatorname{span}(S)\subseteq W$, so $\operatorname{span}(S)\cap W=\operatorname{span}(S)$, i.e., intersection discards the “extraneous” vectors of $W$.
A: 
If $S$ is a subset of a vector space $V$, then $\text{span}(S)$ equals the intersection of all subspaces of $V$ that contain $S$.

This follows directly from the following theorem:
$\textbf{Theorem:}$ 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 the span of $S$.
Let the intersection of all subspaces of $V$ that contain $S$ be denoted by set $U$.
Then, $\; \text{span}(S) \subseteq U \qquad \cdots (1)$
because $\text{span}(S)$ is a subset of all such subspaces of $V$ that contain $S$ (by 2nd part of above theorem), so it must be a subset of their intersection $U$.
Now, what might go unnoticed is that $\text{span}(S)$ is one such subspace, i.e., $\text{span}(S)$ is a subspace of $V$ (by first part of above theorem) $\textit{that contain } S$. Hence, $\text{span}(S)$ must contain the intersection $U$,
i.e., $\; \; U \subseteq \text{span}(S) \qquad \cdots(2)$
From $(1)$ and $(2)$,
$\text{span}(S) = U$
