Is this proof involving the annihilator reasonable?

Question

I'm attempting to prove that if $W$ is a subspace of $V$ and $x \notin W$ that $\exists f \in W^0$ such that $f(x) \neq 0$ where for every subset $S$ of $V$, $S^{0} = \{f \in V^{*} : f(x) = 0 \forall x \in S \}$

My proof is as follows: Suppose $W$ is a subspace of $V$ and that $x \notin W$. $x \in V$, so $x \in V-W$.

Define a functional $f \in V^*$ by $$f(x) = \begin{cases} 0 & x \in W \\ 1 & x \in V - W \end{cases}$$

Then $f \in W^{0}$ because for every subset $S$ of $W$, $x \in S \implies x \in W$, so $f(x) = 0$ and $f(x) \neq 0 \forall x \in V-W$

The answer I have available to me uses a more convoluted argument involving the extension of a basis for $W$. I'm wondering if this is necessary and if my proof is fine as is.

Answer 1

As is pointed in the comments, an extension of a basis for $W$ will be necessary for this, but you can also combine your idea! Suppose that $V$ is finite-dimensional, consider a basis $\beta$ for $W$ and a basis $\alpha$ for $V$ that contains $\beta$. Now define $h : \alpha \to \mathbf F$ by $h(v) = 0$ if $v \in \beta$ and $h(v) = 1$ if $v \in \alpha \setminus \beta$. Then, there is a unique linear extension $f : V \to \mathbf F$ of $h$. It follows that $f$ is your desired linear functional.

Answer 2

If your space $V$ is finite-dimensional, then you can use that $\dim W^0 = \dim V -\dim W$. If $f \in W^0$ implies that $f(x)=0$ then we have $$(W+\operatorname{span}\{x\})^0 = W^0$$ and hence $$\dim V - \dim (W+\operatorname{span}\{x\}) = \dim V-\dim W$$ so the subspaces $W$ and $W+\operatorname{span}\{x\}$ have the same dimension so they must be equal as one contains the other. Therefore $x \in W$ which is a contradiction. We conclude that there is some $f\in W^0$ such that $f(x)\ne0$.