# On annihilator of subspaces.
Let $V$ be a vectorspace over field $F$. Let us denote $V'$ to be the dual space of $V$, namely the set of all linear functionals $V \to F$, equipped with the usual vectorspace structure.
Further, let $U \subset V$ be a subspace of $V$. Define the **annihilator** of $U$ as $U^{0}$ where $$
U^{0} = \set{f \in V':f(x) = 0 \text{ for all } x \in U} = \set{f\in V': U \subset \ker f},
$$the set of all linear functionals on $V$ that annihilates $U$. In other words, $f \in U^{0}$ if and only if $U \subset \ker f$.
We have the following **antitone** property:
>If $U_{1} \subset U_{2}$ are both subspaces of $V$, then $U_{2}^{0} \subset U_{1}^{0}$.
**Proof.** Indeed, take $f \in U_{2}^{0}$, then for all $x \in U_{2}$, we have $f(x)=0$. And for any $x \in U_{1} \subset U_{2}$, we also have $f(x) = 0$. Hence $f \in U_{1}^{0}$.
The question is, what about the converse? If $U_{2}^{0} \subset U_{1}^{0}$, do we necessarily have $U_{1} \subset U_{2}$? This was not immediate obvious to me.
As it turns out, the converse is true if $V$ is finite dimensional, and false if $V$ is infinite dimensional.
To see that it is true when $V$ is finite dimensional, we first have a lemma.
**Lemma.** If $V$ is a finite dimensional vectorspace over $F$, and $U$ some subspace of $V$. Then we have equality $$
U = \set{v \in V : f(v) = 0 \text{ for all } f \in U^{0}}.
$$
**Proof of Lemma.** Let us call right-hand side set as $W$, so $W = \set{v \in V : f(v) = 0 \text{ for all } f \in U^{0}}$. Note we can rephrase $W$ as $$
W = \set{v \in V : v \in \ker f \text{ for all } f \in U^{0}} =
\bigcap_{f \in U^{0}} \ker f
$$Aha, $W$ is just the intersection of all the kernels of $f$ for $f \in U^{0}$.
Now let us establish the equality of $U$ and $W$.
First note $U \subset W$. This is clear because by definition, for any $f \in U^{0}$, we have $U \subset \ker f$. Therefore $U \subset \bigcap_{f \in U^{0}} \ker f$, and hence $U \subset W$.
Next, we show $W \subset U$. Note that if we fix a particular $\phi \in U^{0}$, then yes we have $$
W = \bigcap_{f\in U^{0}} \ker f \subset\ker \phi
$$
So we are done if we can show there is some $\phi \in U^{0}$ such that $\ker \phi = U$ exactly. This is where the finite dimensionality comes in. We construct directly this linear functional $\phi : V \to F$ with $\ker\phi = U$.
Let $u_{1},\ldots, u_{k}$ be a basis for $U$, and extend it to $u_{1},\ldots, u_{k}, v_{1},\ldots, v_{\ell}$ a full basis for $V$. Now define linear functional $\phi : V \to F$ such that $$
\begin{eqnarray}
\phi(u_{1}) &= 0 \\
\vdots \\
\phi(u_{k}) &= 0 \\
\phi(v_{1}) &= 1 \\
\vdots \\
\phi(v_{\ell}) &= 1\\
\end{eqnarray}
$$(Note in a field $F$, we have $0 \neq 1$.) We know this linear map exists as we define it on a basis of $V$, and that by construction $\ker \phi = U$. This is the $\phi \in U^{0}$ that we need.
So, therefore we have $$
W = \bigcap_{f \in U^{0}} \ker f \subset \ker \phi = U
$$and we have $W \subset U$. $\blacksquare$
Great. Now that we have above lemma, we can now show the converse:
> If $V$ is finite dimensional, and $U_{1}$, $U_{2}$ are subspaces of $V$ such that $U_{2}^{0}\subset U_{1}^{0}$. Then we have $U_{1} \subset U_{2}$.
**Proof.** Using the lemma above, let us re-write $U_{1}$ as $$
U_{1} = \bigcap_{f \in U_{1}^{0}} \ker f
$$
Since $U_{2}^{0} \subset U_{1}^{0}$, we have inclusion $$
\bigcap_{f \in U_{1}^{0}} \ker f \subset \bigcap_{f\in U_{2}^{0}} \ker f
$$(Note, this is because taking more intersection makes the resulting set smaller.) Hence we have $U_{1} \subset U_{2}$ as claimed! $\blacksquare$
Now what is this false in infinite dimensions?