Definitions:
Let $V$ be a vector space over $\mathbf{F}$ and $U$ a subset of $V$. The annihilator of $U$, denoted $U^0$ is $$U^0=\{\varphi:V\rightarrow \mathbf{F} \ \ | \ \varphi|_U = 0 \}.$$
Motivation: In general, we know that the annihilator $U^0$ is a subspace of $V$. When $U$ is a subspace of $V$, we also know that $$\text{dim } U^0 = \text{dim } V - \text{dim } U.$$ I'm trying to gain an intuition/ understanding of the annihilator of a general subset. That is why I am asking this question.
Question:
What is the dimension of $U^0$ when $U$ is not a subspace? As a more specific example, what is the dimension of the annihilator of the unit sphere $\mathbf{S} \subset \mathbf{R}^3$ in 3-space?
Idea:
Maybe it should be related to the co-dimension of the smallest subspace of $V$ containing $U$. I know this is easy to determine, and it seems like a linear functional that annihilates one of this subspace should annihilate $U$.
By linearity, if a linear map $L$ vanishes on a vector $v$ (that is, $L(v)=0$) then it has to vanish on the whole line $\Bbb R v$ (called the line "spanned by $v$").
Therefore $U^0=(\operatorname{Span}U)^0$ where $\operatorname{Span}U$ is the smallest linear subspace that contains $U$. From then on you can apply your formula, since $\operatorname{Span}U$ is a linear subspace.
So all you have to do is understand what the subspace spanned by the sphere is.