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

Question

My introductory linear algebra textbook claims that:

If $S$ is a subset of a vector space $V$, then $span(S)$ 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; however, I do not understand what is meant by, "$span(S)$ equals the intersection of all subspaces of $V$ that contain $S$." In addition, it seems to me like this statement is false: $span(S)$ does not necessarily have to equal the intersection of all subspaces of $V$ that contain $S$.

I would greatly appreciate it if someone could 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.

Answer 1

Let $E$ be a subset of $V$ then $L(E)$ is smallest subspace of $V$ which contain $E$.

Answer 2

$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$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.

1. The intersection of any collection of subobjects is a subobject.
2. In particular, the intersection $Y$ of all subobjects of $X$ containing $S$ is a subobject of $X$ containing $S$.
3. If $Z$ is a subobject of $X$ containing $S$, then $Z$ is one of the terms of this intersection, and thus contains $Y$.
4. 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 $S$.

Answer 3

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 $W$.