# The definition of “span” and related theorem.

In wikipedia and the most of the linear algebra texts, The definition of the span is following as.

"Given a vector space V over a field K, the span of a set S of vectors (not necessarily finite) is defined to be the intersection W of all subspaces of V that contain S. W is referred to as the subspace spanned by S, or by the vectors in S. Conversely, S is called a spanning set of W, and we say that S [spans] W."

And, there is a following theorem.

"The span of any subset of given vector space V is subspace of V."

Question: This theorem is true where the subset of V is not finite?

Example) Let $V$ be $R^2$. $~~~W:=\{ (x,y) ~ | ~x ~is ~nonnegative~ \}.$

I think the given theorem is not true where the subset of V is not finite.

The "span" can consider the duplicated vectors.

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The span is an example of a closure. Moreover, as has been pointed out, the intersection of an arbitrary family of vector spaces is a vector space, and so the span of a set is the intersection of all subspaces that contain it. Same thing with closed sets, convex sets, etc. –  lhf Jul 31 '13 at 13:32

Let $V$ be a vector space. The intersection of an arbitrary family of subspaces of $V$ is always a subspace of $V$.

Given $S\subset V$, the span of $S$ is the subspace $$\langle S\rangle=\bigcap_{W\in\cal F}W$$ where $\cal F$ is the family of subspaces of $V$ containing $S$ (note that $\cal F$ is always non-empty as $V\in\cal F$)

This is true regardless of the dimension of $V$ and in fact is a notion independent of that of basis of $V$.

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I'll check the definition again.... –  Chris kim Jul 31 '13 at 12:40

Yes, it is true completely independently of the basis for $V$ or the size of our spanning subset $S$. All you need to do is check the axioms for subspace (or axioms for vector space and show containment). Remember Span$(S)$ is all elements of the form $k_{1}s_{1} + ... k_{n}s_{n}$, where each $k_i\in k$ and each $s_i\in S$. Apparently you are using a different definition of Span, but it is quite easy to show the two are equivalent so I will leave that to you (Hint:they are both the smallest subspace containing S).

$1$. If $x , y\in \text {Span} (S)$, then $x+y \in \text {Span} (S)$

Since $x\in \text {Span} (S)$ then $x=k_{x,1}s_{x,1} + ... k_{x,n}s_{x,n}$ for some $k_{x,i}\in k$ and $s_{x,i}\in S$. Similarly $y=k_{x,1}s_{x,1} + ... k_{x,n}s_{x,n}$, so $x+y=k_{x,1}s_{x,1} + ... k_{x,n}s_{x,n}+k_{x,1}s_{x,1} + ... k_{x,n}s_{x,n}$ so $x+y\in \text{Span} (S)$.

The other axioms are just as straight forward to show so I will leave them to you.

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Hmm.. Ok. I have to check that.... –  Chris kim Jul 31 '13 at 12:31

The intersection of any family of subspaces of a vector space $V$ is a subspace of $V$. If $x,y$ are in the intersection, then $x,y$ are in each member subspace, hence $x+y$ is in each member subspace, hence $x+y$ is in the intersection. The same trick works for scalar multiples and to show that $0$ is in the intersection. So any such intersection of subspaces is a subspace.

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I need to check the definition exactly..... –  Chris kim Jul 31 '13 at 12:39