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."
(http://en.wikipedia.org/wiki/Linear_span)
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.
Best Answer
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.