Let $V$ be a vector space over a field $k$ . Let $S$ be a subset of $V$ . We denote $\Sp(S)$ the span of the set $S$ . Below are some basic properties of spanning sets.

1. If $S\subseteq T$ , then $\Sp(S)\subseteq \Sp(T)$ . In particular, if $\Sp(S)=V$ , every superset of $S$ spans (generates) $V$ .
   Proof. If $v \in \Sp(S)$ , then $v=r_1v_1+\cdots +r_nv_n$ for $v_i\in S$ . But $v_i\in T$ by assumption. So $v\in \Sp(T)$ as well. If $\Sp(S)=V$ , and $S\subseteq T$ , then $V=\Sp(S)\subseteq \Sp(T) \subseteq V$ .
2. If $S$ contains $0$ , then $\Sp(S-\lbrace 0\rbrace)=\Sp(S)$ .
   Proof. Let $T=S-\lbrace 0\rbrace$ . So $\Sp(T)\subseteq \Sp(S)$ by 1 above. If $v\in \Sp(S)$ , then $v=r_1v_1+\cdots + r_nv_n$ . If one of the $v_i$ 's, say $v_i$ , is $0$ , then $v=r_2v_2+\cdots +r_nv_n\in \Sp(T)$ .
3. It is not true that if $S_1\supseteq S_2\supseteq \cdots$ is a chain of subsets, each spanning the same subspace $W$ of $V$ , so does their intersection.
   Proof. Take $V=\mathbb{R}^n$ , the Euclidean space in $n$ dimensions. For each $i=1,2,\ldots$ , let $S_i$ be the closed ball centered at the origin, with radius $1/i$ . Then $\Sp(S_i)=V$ . But the intersection of these $S_i$ 's is just the origin, whose span is itself, not $V$ .
4. $S$ is a basis for $V$ iff $S$ is a minimal spanning set of $V$ . Here, minimal means that any deletion of an element of $S$ is no longer a spanning set of $V$ .
   Proof. If $S$ is a basis for $V$ , then $S$ spans $V$ and $S$ is linearly independent. Let $T$ be the set obtained from $S$ with $v\in S$ deleted. If $T$ spans $V$ , then $v$ can be written as a linear combination of elements in $T$ . But then $S=T\cup \lbrace v\rbrace$ would no longer be linearly independent, contradiction the assumption. Therefore, $S$ is minimal.

   Conversely, suppose $S$ is a minimal spanning set for $V$ . Furthermore, suppose that $S$ is linearly dependent. Let $0=r_1v_1+\cdots r_nv_n$ , with $r_1\ne 0$ . Then \begin{equation} v_1=s_2v_2+\cdots +s_nv_n, \end{equation}where $s_i=-r_i/r_1$ . So any linear combination of elements in $S$ involving $v_1$ can be replaced by a linear combination not involving $v_1$ through equation (1). Therefore $\Sp(S)=\Sp(S-\lbrace v\rbrace)$ . But this means that $S$ is not minimal, contrary to our assumption. Therefore, $S$ must be linearly independent.

Remark. All of the properties above can be generalized to modules over rings, except the last one, where the implication is only one-sided: basis implying minimal spanning set.