Exercise 1.4.13

We prove $\mathrm{span}(S_1\cup S_2)\subset \mathrm{span}(S_1)+\mathrm{span}(S_2)$ first. For $v\in \mathrm{span}(S_1\cup S_2)$ we have $v={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^n{a}_i{x}_i+{\sum }_{j=1}^m{b}_j{y}_j$ with $x_i\in S_1$ and $y_j\in S_2$. Since the first summation is in $\mathrm{span}(S_1)$ and the second summation is in $\mathrm{span}(S_2)$, we have $v\in \mathrm{span}(S_1)+\mathrm{span}(S_2)$. For the converse, let $u+v\in \mathrm{span}(S_1)+\mathrm{span}(S_2)$ with $u\in \mathrm{span}(S_1)$ and $v\in \mathrm{span}(S_2)$. We can right $u+v={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^n{a}_i{x}_i+{\sum }_{j=1}^m{b}_j{y}_j$ with $x_i\in S_1$ and $y_j\in S_2$ and this means $u+v\in \mathrm{span}(S_1\cup S_2)$.