# Span of subsets and span of their union.

Let $$S$$ and $$T$$ be two subsets of a vector space $$E$$. I want to prove

$$span(S) + span(T) = span(S \cup T)$$.

Here is my approach: let $$span(S) = span(v_1, \dots, v_m)$$ and $$span(T) = span(u_1, \dots, u_n)$$.

• $$span(S) + span(T) \subseteq span(S \cup T)$$. Let $$w \in span(S) + span(T)$$. Then $$w = x+y$$, where $$x \in span(S)$$ and $$y \in span(T)$$. Since $$x \in span(S)$$, $$x = \alpha_1 v_1 + \dots + \alpha_m v_m$$ for some $$\alpha_i \in \mathbb{R}$$, $$i=1, \dots, m$$. Similarly, $$y = \beta_1 u_1 + \dots + \beta_n u_n$$ for some $$\beta_i \in \mathbb{R}$$, $$i = 1, \dots, n$$. Therefore, since $$w = \alpha_1 v_1 + \dots + \alpha_m v_m + \beta_1 u_1 + \dots + \beta_n u_n$$, $$w \in span(v_1, \dots, v_m, u_1, \dots, u_n$$) i.e. (not sure if the following step is right) $$w \in span(S\cup T)$$.
• $$span(S \cup T)\subseteq span(S) + span(T)$$. Let $$w \in span(S \cup T)$$, then $$w = \alpha_1 v_1 + \dots + \alpha_m v_m + \beta_1 u_1 + \dots + \beta_m u_m = x+y$$, where $$x \in span(S)$$ and $$y \in span(T)$$. Therefore, $$w \in span(S) + span(T)$$.

Im not sure if those steps are right, mainly the ones involving the fact that a vector belongs to the union of subsets. Is this correct or is there another way, more elegant maybe, to prove this? Thank you!

That is not correct. You are assuming that $$S$$ and $$T$$ are finite, but that is not part of the assumptions of the problem.
Since $$S\subset S\cup T$$, $$\operatorname{span}(S)\subset\operatorname{span}(S\cup T)$$. For the same reason, $$\operatorname{span}(T)\subset\operatorname{span}(S\cup T)$$, and therefore$$\operatorname{span}(S)+\operatorname{span}(T)\subset\operatorname{span}(S\cup T).$$
On the other hand, if $$v\in\operatorname{span}(S\cup T)$$, then there are vectors $$v_1,\ldots,v_k\in S$$, there are vectors $$v_{k+1},\ldots,v_l\in T$$ and there are scalars $$\lambda_1,\ldots,\lambda_l$$ such that$$v=\overbrace{\lambda_1v_1+\cdots+\lambda_lv_k}^{\phantom{\operatorname{span}(S)}\in\operatorname{span}(S)}+\overbrace{\lambda_{k+1}v_{k+1}+\cdots+\lambda_lv_l}^{\phantom{\operatorname{span}(T)}\in\operatorname{span}(T)},$$and therefore $$v\in\operatorname{span}(S)+\operatorname{span}(T)$$.
• By the way, just to clarify myself. $span(S) + span(T) \subset span(S\cup T)$ because $span(S\cup T)$ is a subspace and therefore closed under addition, is this reasoning correct? Feb 12, 2021 at 9:27