91Ó°ÊÓ

(a) To show that if \(S \subseteq T\), then \(\operatorname{span}(S) \subseteq \operatorname{span}(T)\), consider an arbitrary element \(x \in \operatorname{span}(S)\). Then, \(x\) can be expressed as a linear combination of elements in \(S\). Since \(S \subseteq T\), these elements are also in \(T\), and \(x\) can be expressed as a linear combination of elements in \(T\). This implies that \(\operatorname{span}(S) \subseteq \operatorname{span}(T)\). (b) To show that \(\operatorname{span}[\operatorname{span}(S)]=\operatorname{span}(S)\), first establish that \(\operatorname{span}(S) \subseteq \operatorname{span}[\operatorname{span}(S)]\). Then, apply the result from part (a) to the sets \(\operatorname{span}(S)\) and \(\operatorname{span}[\operatorname{span}(S)]\) to show that \(\operatorname{span}[\operatorname{span}(S)] \subseteq \operatorname{span}(S)\). Combining these two results, we can conclude that \(\operatorname{span}[\operatorname{span}(S)]=\operatorname{span}(S)\).