If $E$ is a subset of $V^*$, then the annihilator $E^0$ is (technically) a subset of $V^{**}$. If we choose to identify $V$ and $V^{**}$ as in Theorem 17, then $E^0$ is a subspace of $V$, namely, the set of all $\alpha$ in $V$ such that $f(\alpha)=0$ for all $f$ in $E$. In a corollary of Theorem 16 we noted that each subspace $W$ is determined by its annihilator $W^0$. How is it determined? The answer is that $W$ is the subspace annihilated by all $f$ in $W^0$, that is, the intersection of the null spaces of all $f$’s in $W^0$. In our present notation for annihilators, the answer may be phrased very simply : $W = (W^0)^0$.
Theorem 18. If $S$ is any subset of a finite-dimensional vector space $V$, then $(S^0)^0$ is the subspace spanned by $S$.
Proof. Let $W$ be the subspace spanned by $S$. Clearly $W^0 = S^0$. Therefore, what we are to prove is that $W = W^{00}$. We have given one proof. Here is another. By Theorem 16, $\mathrm{dim}(W)+ \mathrm{dim}(W^0)$ $= \mathrm{dim}(V)$ and $\mathrm{dim}(W^0)+ \mathrm{dim}(W^{00})$ $=\mathrm{dim}(V^*)$. Since $\mathrm{dim}(V)$ $= \mathrm{dim}(V^*)$, we have $\mathrm{dim}(W)= \mathrm{dim}(W^{00})$. Since $W$ is a subspace of $W^{00}$, we see that $W=W^{00}$.
I find this theorem vague mainly because we’re trying show two sets are equal but they aren’t equal. Is this theorem “important”? Is it okay to skip it?
My attempt: Hoffman’s didn’t show lots of details in proof. I will try to fill those details. (1) Claim: $E^0=\{\alpha\in V|f(\alpha)=0\text{, }\forall f\in E\}$. Proof: By definition of annihilator, $E^0=\{L\in V^{**}|L(f)=0\text{, }\forall f\in E\}$. By theorem 17 section 3.6, $\forall L\in E^0$, $\exists !\alpha \in V$ such that $\phi(\alpha)=L_\alpha=L$. Thus $E^0$ $=\{L\in V^{**}|L(f)=0\text{, }\forall f\in E\}$ $= \{L_\alpha\in V^{**}|L_\alpha(f)=f(\alpha)=0\text{, }\forall f\in E\}$ $=\{\alpha \in V| f(\alpha)=0\text{, }\forall f\in E\}$. Hence $E^0= \{\alpha \in V| f(\alpha)=0\text{, }\forall f\in E\}$. (2) Claim: $W=\bigcap_{f\in W^0}N_f$. Proof: By theorem 16 corollary 1 section 3.5, $W=\bigcap_{k\lt i\leq n}N_{f_i}$. By elementary set theory, $W =\bigcap_{k\lt i\leq n}N_{f_i}\supseteq \bigcap_{f\in W^0}N_f$. Conversely, $W\subseteq N_f$, $\forall f\in W^0$. Thus $W\subseteq \bigcap_{f\in W^0}N_f$. Hence $W= \bigcap_{f\in W^0}N_f$ $=\{\alpha\in V|\alpha\in N_f\text{, }\forall f\in W^0\}$ $=\{\alpha\in V|f(\alpha)=0\text{, }\forall f\in W^0\}$. So by claim $(1)$, $W$ $= \{\alpha\in V|f(\alpha)=0\text{, }\forall f\in W^0\}$ $=(W^0)^0$ $=W^{00}$. (3) Claim: $(\mathrm{span}(S))^0=S^0$. Proof: Let $f\in (\mathrm{span}(S))^0$. Then $f(x)=0_F$, $\forall x\in \mathrm{span}(S)$. Since $S\subseteq \mathrm{span}(S)$, we have $f(x)=0_F$, $\forall x\in S$. Thus $f\in S^0$. Hence $(\mathrm{span}(S))^0\subseteq S^0$. Conversely suppose $f\in S^0$. Then $f(x)=0_F$, $\forall x\in S$. let $v\in \mathrm{span}(S)$. So $v=\sum_{i\in J_n}a_i\cdot u_i$, where $n\in \Bbb{N}$, $a_i\in F$, $u_i\in S$. Since $f$ is a linear map, precisely $f\in V^*$$=L(V,F)$, we have $f(v)$ $=f(\sum_{i\in J_n}a_i\cdot u_i)$ $= \sum_{i\in J_n}a_i\cdot f(u_i)$ $= \sum_{i\in J_n}a_i\cdot 0_F$ $=0_F$. Since $v$ was arbitrary, $f(v)=0_F$, $\forall v\in \mathrm{span}(S)$. Thus $f\in (\mathrm{span}(S))^0$. Hence $S^0\subseteq (\mathrm{span}(S))^0$. So $S^0= (\mathrm{span}(S))^0$. Which implies $(S^0)^0$ $= ((\mathrm{span}(S))^0)^0$ $=S^{00}$ $= (\mathrm{span}(S))^{00}$ $=\mathrm{span}(S)$. Thus $S^{00}=\mathrm{span}(S)$. Is my proofs correct?
Question: “Since $W$ is a subspace of $W^{00}$….”. How $W$ is a subspace of $W^{00}$? It seems absurd!