Let $\text{span}(S) = \lbrace v \in V \mid v\ \text{is a linear combination of vectors in}\ S\rbrace$.
I need to show that $S$ is contained within $\text{span}(S)$. I know if $S$ is nonempty, $0$ is an element of $\text{span}(S)$. Is there any way to build upon that fact in order to prove that $S$ is contained within $\text{span}(S)$? Should we show that $\text{span}(S)$ is a subset of $V$ in order to say that $S$ is contained in $\text{span}(S)$?
Definition 2.1 (Span). Let $V$ be a vector space over a field $F$. Let $S \subseteq V$ be any subset of vectors. The span of $S$, denoted $\text{span}(S)$ is the set of all linear combinations of vectors in $S$. That is
$$
\text{span}(S) := \{v \in V \mid v\ \text{is a linear combination of vectors in}\ S\}\\
= \{v \in V \mid \exists c_1, \ldots, c_m \in F,\ v_1, \ldots v_m \in S,\ \text{s.t.}\ v = c_1 v1 + \ldots + c_m v_m \}
$$
Can something similar to this answer the above question?
Let $W$ be a subspace containing $S$. We need to show that $W$ contains
$\text{span}(S)$. This is obvious. Since $W$ is a subspace, it is closed under
addition and scalar multiplication. Therefore, if $W$ contains $S$, it must
contain every linear combinations of elements of $S$. But the set of all
possible linear combinations of elements of $S$ is precisely $\text{span}(S)$.
Best Answer
Any element $ s \in S $ is trivially a linear combination of elements from $ S $, since, obviously $ s = 1*s $.
You can imagine span(S) as the set obtained by taking elements of S and "putting them together" in every possible way. Any vector from S can be obtained if you just take it and no other vectors. So it's in span(S) and $ S \subset span(S) $.