This is a theorem from Roman's textbook "Advanced Linear Algebra"(p.$48$).

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> **Theorem** $1.9.$ Let $V$ be a nonzero vector space. Let $I$ be a linearly independent subset of $V$ and let $S$ be a spanning set of $V$ containing $I$. Then there is a basis $B$ such that $I \subseteq B \subseteq S$. 
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**Proof:**

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> Let $A = \{$all linear independent subsets of $S$ containing $I \}, A \neq \varnothing$ since $I \in A$
>  Now, let $C = \{I\_k | k \in K \}$ be any chain in a poset $A$. 
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> $\bigcup\limits\_{k \in K} I\_k \in A$ since it is linearly independent and is contained in $S$. So $\bigcup\limits\_{k \in K} I\_k$ is an upper bound for $C$, that is, any chain in $A$ has an upper bound, so Zorn's lemma tells us that $A$ contains a maximal element $B$.
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> Now we need to prove that that $B$ is a basis for $V = span(S)$. 
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> If any $s \in S$ is not a linear combination of the elements of $B$, then $\{s \} \cup B \subseteq S$ is also linearly independent contradicting the maximality of $B$. So, $S \subseteq span(B)$ **and so $V = span(S) \subseteq span(B)$**.
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What is don't understand is why "and so"?
I got that $S \subseteq span(B)$. How does is tell us that $V = span(S) \subseteq span(B)$?