$V$ is $K$-Vector Space and $0<n \in \Bbb{N}$, I define with $(v_1,...,v_n) \in V^n$ and $T \subseteq V$:
$$\mathscr{L}((v_1,...,v_n)):=\{x|\exists (\alpha_1,...,\alpha_n)\in K^n:x=\sum_{i\in J_m:=\{1,...,n\}}\alpha_i\, v_i\}$$$$<T>:=\operatorname{Span}(T):=\{x|\exists (w_1,...,w_n)\in T^n: x \in \mathscr{L}((w_1,...,w_n))\}$$
Is it correct?
On
Another way to put it, which I prefer because it can easily be generalized, is the following : since the set of subspaces of $V$ is closed under arbitrary intersections, then there is a least (w.r.t. inclusion) subspace containing $T$. We let $<T>$ or $Span(T)$ denote this subspace.
On
Let be $m \in \Bbb{N}=\big \{1,2,3,...,n,n+1,...\big \}$, we can define $$J_m:=\big \{x \in \Bbb{N}\big | x \leq m \big \}=\big \{1,2,3,...,m \big \}$$ Let be $T \subseteq V$ and $V$ a $K$-vector Space, we can define $$\operatorname{Span}\big (T \big):=\mathscr{L}\big (T\big ):=$$$$:= \begin{cases}\big \{0_V \big \} & \text{ if } \,\,\,T=\emptyset \\ \bigg \{z\big| \exists m\in \Bbb{N}:\bigg(\exists v \in V^{J_m}, \exists\alpha \in K^{J_m}:\bigg( z= \displaystyle\sum_{i=1}^m \alpha_i\, v_i\bigg)\bigg)\bigg\} & \text{ otherwise}\end{cases}$$ $$\big \langle T \big \rangle := \bigcap \big \{x\subseteq V| T \subseteq x \wedge x \text{ is subspace von } V\big \} $$ Now you can prove that $\big \langle T \big \rangle=\operatorname{Span}\big (T \big ) $
This looks correct, though a little overboard, to me. There are more succinct ways to write this, namely "the span of $T$ is the set of all vectors that are linear combinations of elements of $T$"