The Creation of Unions is Equal to the Sum of Creations
Theorem1
Let $S_{1}, S_{2}$ be a subset of the vector space $V$. Then, the following holds.
$$ \span(S_{1} \cup S_{2}) = \span(S_{1}) + \span(S_{2}) $$
Here, $\span$ means generation, and $+$ means the sum of sets.
Proof
$\span(S_{1} \cup S_{2}) \subset \span(S_{1}) + \span(S_{2})$
Let $v \in \span(S_{1} \cup S_{2})$. Then, $v$ can be expressed as follows: $$ v = \sum\limits_{i=1}^{n}a_{i}x_{i} + \sum\limits_{j=1}^{m}b_{j}y_{j},\quad x_{i}\in S_{1},\ y_{j} \in S_{2} $$ The first sum belongs to $\span(S_{1})$, and the second sum belongs to $\span(S_{2})$. Therefore, $v \in \span(S_{1}) + \span(S_{2})$.
$\span(S_{1} \cup S_{2}) \supset \span(S_{1}) + \span(S_{2})$
$$ u = \sum\limits_{i=1}^{n}a_{i}x_{i} \quad \text{and} \quad v = \sum\limits_{j=1}^{m}b_{j}y_{j},\quad x_{i} \in S_{1},\ y_{j} \in S_{2} $$ Regarding such $u \in \span(S_{1}), v \in \span(S_{2})$, let $u + v \in \span(S_{1}) + \span(S_{2})$. Then, since $u + v = \sum\limits_{i=1}^{n}a_{i}x_{i} + \sum\limits_{j=1}^{m}b_{j}y_{j}$ holds, $u + v \in \span(S_{1} \cup S_{2})$.
■
Stephen H. Friedberg, Linear Algebra (4th Edition, 2002), p34 ↩︎