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Source. Lemma 2, printed p. 151 (published PDF).

Statement. If $I$ is independent and $I\subseteq A$, there is a unique largest set $S$ such that

$$I\subseteq S\subseteq A,\qquad r(S)=|I|.$$

It is

$$\operatorname{sp}_A(I) =I\cup\{e\in A\setminus I:I\cup\{e\}\text{ is dependent}\}.$$

Proof. At least one eligible set exists, namely $I$. Since $A$ is finite, choose an inclusion-maximal eligible set $S$. If $I\cup\{e\}$ is independent for $e\in A\setminus I$, then $e$ cannot lie in $S$, since it would give an independent subset of $S$ larger than $r(S)=|I|$.

If instead $I\cup\{e\}$ is dependent, no element of $(S\cup\{e\})\setminus I$ can be added to $I$: elements of $S$ cannot enlarge an independent set already of size $r(S)$, and $e$ cannot by assumption. Hence $I$ is maximal independent in $S\cup\{e\}$, so this union has rank $|I|$. Maximality of $S$ forces $e\in S$. These two observations identify $S$ with the displayed set. Every eligible set is contained in it, proving both uniqueness and the largest-set assertion.

Consequence used in the exchange proof. If $S=\operatorname{sp}_A(I)$ and $I'\subseteq S$ is independent of size $|I|$, then $\operatorname{sp}_A(I')=S$. Indeed, adjoining an element of $S$ to $I'$ cannot increase its independent size, so $S\subseteq\operatorname{sp}_A(I')$. The latter set has rank $|I|$ and contains $I$, so the largest-set assertion for $I$ gives the opposite inclusion. This argument is part of the same span deduction. $\square$