Annihilator of a vector space $V$ is the zero subspace of $V^*$

Question

I am reading Hoffman and Kunze's Linear Algebra and in Section 3.5, page 101, they define the annihilator of a subset as follows:

Definition. If $V$ is a vector space over the field $F$ and $S$ is a subset of $V$, the annihilator of $S$ is the set $S^0$ of linear functionals $f$ on $V$ such that $f(\alpha) = 0$ for every $\alpha$ in $S$.

In the following paragraph, they say that

If $S = V$, then $S^0$ is the zero subspace of $V^*$. (This is easy to see when $V$ is finite-dimensional.)

My question is regarding the statement within the parentheses. Is there some subtlety when considering infinite-dimensional vector spaces? Below is my proof of the fact that $S^0 = \{ 0 \}$ when $S = V$.

Let $S = V$. If $f \in V^*$ is a non-zero functional, then $f(v) \neq 0$ for some $v \in V = S$. So, $f \not\in S^0$. So, $S^0 \subseteq \{ 0 \}$. Also, if $0$ is the zero functional, then it maps every $v \in V = S$ to $0$, and so $\{ 0 \} \subseteq S^0$. Hence, $S^0$ is the zero subspace of $V^*$.

I don't see where the dimension of $V$ plays a role in the above proof. Is the statement given within the parantheses redundant, or am I missing something crucial in understanding the case $\dim V = \infty$?

Answer 1

Your argument is correct since we have that $$ f \in V^0 \iff \forall v \in V: f(v) = 0 \iff f = 0. $$ (This is just a compressed version of your argumentation.)

What also holds is a related statement, namely that the set $$ U := \{v \in V \mid \forall f \in V^*: f(v) = 0\} $$ is just the zero vector space: We can extend every $v \in V$, $v \neq 0$ to a basis $\mathcal{B}$ of $V$, and then define a (unique) linear function $f \colon V \to F$ with $f(v) = 1$ and $f(v') = 0$ for all $v' \in \mathcal{B}$, $v' \neq v$ (the values $f(v')$ do not matter). Then $f(v) = 1 \neq 0$, so $v \notin U$.

Because this construction uses a basis $\mathcal{B}$ of $V$, one can argue that there is a difference between the finite-dimensional and infinite-dimensional case for this related statement. (I don’t know if Hoffman & Kunze prove the existence of bases for infinite dimensional vector spaces.) My best guess is that Hoffman & Kunze mixed this up when writing the quoted text.