I have these two definitions of span:
Span:
Suppose a vector space $(V,+,\cdot)$, and
$$S = \{u_1,\cdots,u_n\}$$
(and $S$ is a subset of $V$, not a subspace)
$$[S]=:\cap_{w\subset V, w\supseteq S} W$$
In other words, $[S]$ is, by definition, the intersection of all $W$, such that $W$ is a subspace of $V$ and $W$ contains $S$.
Span: $$S = \{u_1,\cdots,u_n\}$$ The span of $S$, denote by $[S]$ is, then: $$W = \{\alpha_1u_1+\cdots+\alpha_nu_n|\alpha_i\in\mathbb R\}$$
I need to prove that if $U = [S_1]$ is a subset of $V$ and $W = [S_2]$ is also a subset of $V$, then $$[S_1]+[S_2] = [S_1\cup S_2]$$
where, I think, $[S_1]+[S_2]$ means sum of subspaces
By my second definition, I did:
$$S_1 = {u_1,\cdots,u_n} \implies [S_1] = \{\lambda_1 u_1+\cdots+\lambda_n u_n|\lambda_i\in\mathbb R\}$$ $$S_2 = {w_1,\cdots,w_n} \implies [S_2] = \{\gamma_1 w_1+\cdots+\gamma_n w_n|\gamma_i\in\mathbb R\}$$ Then, the sum of two subspaces $U$ and $W$ is:
$$\{u+w|u\in U, w\in W\}$$ therefore
$$[S_1] + [S_2] = \{\lambda_1 u_1+\cdots+\lambda_n u_n + \gamma_1 w_1+\cdots+\gamma_n w_n|\lambda_i,\gamma_i \in \mathbb R\}\tag{1}$$
And note that: $$S_1\cup S_2 = \{u_1,\cdots,u_n,w_1,\cdots,w_m\}$$
therefore:
$$[S_1\cup S_2] = \{\beta_1 u_1+\cdots+\beta_n u_n+\beta_{n+1} w_1+\cdots+\beta_{n+m}w_m|\beta_i\in\mathbb R\}\tag{2}$$
So $(1)$ and $(2)$ are essentially the same thing. Is my reasoning correct? I did not prove that one set is a subset of the other, but I worked on the two sides of the equations and found what is essentially the same set. I think there's no problem with it, right?
Update: here's my idea for a better proof using the first definition:
$$S_1 = \{v_1,\cdots,v_n\}\\S_2 = \{u_1,\cdots,u_m\}\\ S_1 \cup S_2 = \{v_1,\cdots,v_n,u_1,\cdots,u_m\}\\ k\in \operatorname{span}(S_1\cup S_2)\implies \\k = a_1v_1+\cdots+a_nv_n + b_1u_1+\cdots+b_mu_m \\= (a_1v_1+\cdots+a_nv_n)+(b_1u_1+\cdots+b_mu_m) \in \operatorname{span}(S_1)+\operatorname{span}(S_2)$$ $$k\in \operatorname{span}(S_1)+\operatorname{span}(S_2)\implies\\k = (a_1v_1+\cdots+a_n v_n) + (b_1u_1 + \cdots+b_mu_m) = \\ a_1v_1+\cdots+a_nv_n + b_1u_1+\cdots+b_mu_m \in \operatorname{span}(S_1 \cup S_2)$$ $k$ was chosen arbitrarily, therefore everything above Works for all $k$. Then we have: $$\operatorname{span}(S_1) + \operatorname{span}(S_1) \subseteq \operatorname{span}(S_1\cup S_2)\\\operatorname{span}(S_1\cup S_2)\subseteq \operatorname{span}(S_1) + \operatorname{span}(S_1)\\ \implies\\\operatorname{span}(S_1) + \operatorname{span}(S_1) = \operatorname{span}(S_1\cup S_2)$$
The main question is: how to prove the same thing, but for infinite sets $S_1$ and $S_2$? This is a proof that might envolve the first definition, because it works for infinite sets. Of course it has something to do with proving $[S_1]+[S_2]\subseteq [S_1 \cup S_2]$ and $[S_1\cup S_2] \subseteq [S_1]+[S_2]$, but how to proceed?
What I tried:
I know by definition that
$$[S_1\cup S_2] = \cap_{w\subset V, w\supseteq S_1\cup S_2} W$$
Update:
To illustrate, I've made this figure:
that shows some possible $W_i$ of the intersection on the definition.
My idea is: pick any $x\in [S_1\cup S_2]$, therefore this $x$ is in the purple intersection of the figure. If it is in the intersection, it is in all subspaces $W_i$. But since $W_i$ is a subspace that contains $S_1\cup S_2$, one of these subspaces can be $[S_1]+[S_2]$ because:
$$[S_1]+[S_2] = \{u+v| u\in [S_1], v \in [S_2]\}$$ And, remember, this is a subspace, but must also contain $S_1\cup S_2$ by the definition. But it contains, because we could show that there exists $2$ subsets of $[S_1]+[S_2]$ in the form:
$$w_1 = \{u+0|u\in [S_1], 0 \in [S_2]\} = [S_1] \supset S_1 $$ $$w_2 = \{0+v|0\in [S_1], v \in [S_2]\} = [S_2] \supset S_2 $$
Since both $w_1\supset S_1$ and $w_2\supset S_2$ are in $[S_1]+[S_2]$, then $w_1\cup w_2 = [S_1]\cup [S_2] \supset S_1\cup S_2$
I showed that such subspace $[S_1]+[S_2]$ matches the definition, so all $x\in [S_1\cup S_2]\subset [S_1]+[S_2]$.
I think, then, one side is proved. I need to prove the converse.
