prove that $Span(A)$ is a subspace of $Span(B)$ Let $A$ and $B$ be two subsets of a vector space $V$ and assume that
$A \subseteq B$. Prove that $Span(A) \le Span(B)$.
What I tried: to prove that $Span(A)$ is a subset of $Span(B)$, I tried to use the definition of spanning sets. I wasn't sure how to prove that + and * are closed for $Span(B)$.
 A: Let me prove that $\text{span}(A)\leq \text{span}(B)$, so take a vector $v\in\text{span}(A)$. By definition of the span, there exist vectors $a_1,\ldots,a_n$ in $A$, and scalars $c_1,\ldots,c_n$ such that
$$v = c_1a_1+\cdots+c_na_n.$$
Since $A\subseteq B$, the elements $a_1,\ldots,a_n$ will be part of the set $B$, that is, $a_i\in B$ for all $1\leq i\leq n$. But then $v\in\text{span}(B)$, since $v$ is a linear combination of vectors from $B$. This shows that $\text{span}(A)\leq\text{span}(B)$.
A: Depending on your definition of the span you have
\begin{align*}
 \mathrm{span}(A)
 &=
 \left\{
  \sum_{i=1}^n \lambda_i a_i
  \,\middle|\,
  n \in \mathbb{N}, \lambda_i \in K, a_i \in A
 \right\} \\
&\subseteq
 \left\{
  \sum_{i=1}^n \lambda_i b_i
  \,\middle|\,
  n \in \mathbb{N}, \lambda_i \in K, b_i \in B
 \right\}
= \mathrm{span}(B)
\end{align*}
or 
\begin{align*}
 \mathrm{span}(A)
 &= \bigcap\{U \subseteq V \mid\text{$U$ is subspace, $A \subseteq U$} \} \\
 &\subseteq \bigcap\{U \subseteq V \mid\text{$U$ is subspace, $B \subseteq U$} \}
 = \mathrm{span}(B).
\end{align*}
A: How does the members of $Span(A)$ look like? 
If $x\in Span(A)$ then $x$ is a linear combination of members of $A$. Do this linear combination exist in $Span(B)$?
To elaborate if $V$ is above some field $\mathbb F$. then a linear combination $x$ in $V$ is of the form $$x=\sum_{i=1}^{n}c_i v_i$$for $c_i \in \mathbb F$ and $v_i \in V$. then if we say that $x$ is a linear combination of members of $A$ it means that $v_i \in A$ for all $1\le i\le n$.
A: If $v\in\text{span}(A)$ then we can write $v=\lambda_1a_1+\dots+\lambda_na_n$ for $\lambda_i\in F$ and $a_i\in A$. However, each $A\subseteq B$ so $a_i\in B$ for each $i$, and hence $v=\lambda_1a_1+\dots+\lambda_na_n\in\text{span}(B)$
