Showing the existance of a base with certain conditions Let $V$ be a vector space and $S \subset T \subset V$, with $S$ being an linearly independent set of vectors and $T$ being a generator set of $V$.
Show that exists an base $B$ for $V$ with $S \subset B \subset T$.
My thoughts are two:

*

*If $S$ is a linear independent subset of $T$, with $T$ being an generator set, then $S$ is a base for some subspace $V' \subset V$. Then we can extend $S$ to be a base $B$ for $V$. But I can't think in how can I extend $S$ to have $B \subset T$ guaranteed.

*If $T=\{w_1,...,w_m\}$ generates $V$, then $m \geq n$, with $n$ being the dimension of $V$. So the vectors of $T$ can't be linearly independent, and we can eliminate some $w_i$ until we get an linearly independent subset of $T$ that generates $V$, that being the base $B$. But how can $I$ do that process ensuring that $S \subset B$? I know that $S$ is an linearly independent set, but in that process, some vectors of $S$ could be lost, right?

I can't conclude the question because of that two doubts. Any leads? Thanks!
 A: A bit late to the party, but here we go: The answer to this question depends on whether $T$ is finite or infinite. But in principle the first idea you presented already goes in the right direction.
If $T=\{w_1,\ldots,w_n\}$ is finite, one usually starts with
$$
B:=\begin{cases}
S&\text{if }w_1\in\operatorname{span}(S)\\
S\cup\{w_1\}&\text{else.}
\end{cases}
$$
The $j$-th step ($j=2,\ldots,n$) then consists of keeping $B$ as it is if $w_j\in\operatorname{span}(B)$, and adding $w_j$ to $B$ otherwise. After $n$ steps one is left with a set $B$ which is linearly independent, and satisfies $S\subseteq B\subseteq T$ as well as $T\subseteq\operatorname{span}(B)$. But the latter implies $V=\operatorname{span}(T)\subseteq\operatorname{span}(\operatorname{span}(B))=\operatorname{span}(B)$ meaning $B$ is a basis of $V$ as claimed.
Now if $V$ is of infinite dimension then this result still holds, but the proof we presented above breaks down because our selection process may not terminate as it could be infinite. This is where Zorn's lemma comes into play; the argument orients itself towards the proof that every vector space has a basis. Define
$$
P:=\{B\subseteq T:B\text{ linearly independent, }S\subseteq B\}\,.
$$
This set is non-empty ($S\in P$) and partially ordered by means of the set inclusion, hence it has a maximal element $B$, i.e. if $B'\in P$ satisfies $B\subseteq B'$ then $B$ and $B'$ have to coïncide. To show that $B$ is a basis of $V$ we have to show that $B$ is linearly independent (holds by definition of $P$) and that $\operatorname{span}(B)=V$.
Indeed assume that $\operatorname{span}(B)\subsetneq V$ so there exists $x\in V\setminus\operatorname{span}(B)$. By assumption $x\in\operatorname{span}(T)=V$ so $x=\sum_{i=1}^mc_iw_{\alpha_i}$ for some $m\in\mathbb N$, $c_i\in\mathbb K$, $w_{\alpha_i}\in T$. But this means that $w_{\alpha_i}\not\in\operatorname{span}(B)$ for some $i=1,\ldots,m$ so $B\cup\{w_{\alpha_i}\}$ is linearly independent; therefore $B\cup\{w_{\alpha_i}\}\in P$. This contradicts maximality of $B$, hence $\operatorname{span}(B)=V$ and $B$ is a basis of $V$ as claimed.
