Unnecessary assumption in Roman's Advanced Linear Algebra, exercise 3.10? The following is an exercise(3rd edition, pg. 107, no. 10) from Roman's Advanced Linear Algebra:
Let $S$ be a proper subspace of a finite-dimensional vector space $V$ and let $v \in V \backslash S $. Show that there is a linear functional $ f \in V^* $ for which $f(v) = 1$ and $f(s) = 0$ for all $s \in S$.
My idea was to take the complement of $S$ to be $T$ so that $S \oplus T = V$, and then let $v = a + b$ for $a \in S, b \in T$. Because $v \not\in S$, $ b \neq 0$, and I would be able to extend $\{b\}$ to a basis of $V$, and then produce a functional by assigning $1$ to $b$ and $0$ to every other vector in the basis.
The problem is that I cannot seem to find the part of the proof that requires finite dimensionality. I would greatly appreciate it if you could identify where finite dimentionality is implicitly used in the above argument, or if the above argument has an error, or if the author assumed unnecessary conditions.
 A: There is no such thing as the complementary subspace to a subspace; there are in general many of them. If you assume the axiom of choice, for instance under the guise of the statement that any independent family of vectors can be completed to a basis of the space, then you can find such a complementary space (proving existence of at least one of them) as follows. Choose a basis of the subspace (which is possible by mentioned assumption), and complete it to a basis (again possible or the same reason); the vectors in the second part of the basis span a complementary subspace. The assumption of finite dimensionality was probably made so that you can avoid invoking the axiom of choice (sometimes assumptions are not strictly necessary, but simply made to simplify the task of proving the statement).
You don't need to mention a complementary subspace explicitly though. Just take a basis of $S$, add $v$ (which is independent of it). complete to a basis, and take the linear functional that takes the value $0$ on all basis vectors except $v$ where it takes the value$~1$. (In other words take the dual basis vector corresponding to $v$ for the basis.)
