The subspace sum of a point and a closed subspace is closed In projective geometry, a polarity is a map $\ell\mapsto\ell^\perp$ on the subspaces of $\Bbb P$ satisfying the axioms:


*

*$\Bbb P^\perp=0$

*$\ell\subseteq m\implies m^\perp\subseteq\ell^\perp$

*If $P$ is a point, then $P^\perp$ is a hyperplane (i.e. there is a point $Q\notin P^\perp$ with $Q+P^\perp=\Bbb P$) and $P^{\perp\perp}=P$.


A subspace $\ell$ is called closed if $\ell^{\perp\perp}=\ell$. (Note that this usage of "closed" may not have anything to do with the topological meaning, and ${}^\perp$ may not be an orthogonal complement.)
I am stuck trying to show that if $\ell$ is closed and $P$ is a point, then the subspace sum $\ell+P$ is closed.
Suppose $\ell+P\subsetneq \ell+P+Q\subseteq(\ell+P)^{\perp\perp}$ for some point $Q$. Then $(\ell+P)^\perp=(\ell+P+Q)^\perp$, but this doesn't seem to lead anywhere else fruitful. This can also be written as $m\cap P^\perp=m\cap(P+Q)^\perp$ where $m=\ell^\perp$.
Note that this question is similar, but involves the topological/limit definition of closed, whereas here I have only the axioms of this wannabe orthogonal projection operator available.
These statements are used without proof in this paper, on p. 10.
 A: There are various facts to check first.


*

*Clearly $\ell\subset \ell^{⊥⊥}$ for all subspaces.

*Moreover we also have for all subspaces $\ell_1,\ell_2$ the equality
$$
(\ell_1+\ell_2)^⊥= \ell_1^⊥\cap \ell_2^⊥.
$$
In fact, clearly $(\ell_1+\ell_2)^⊥\subset \ell_1^⊥\cap \ell_2^⊥$, so assume that $P\in  \ell_1^⊥\cap \ell_2^⊥$. Then 
$$
P\in \ell_1^{\bot}\quad \Rightarrow\quad \ell_1\subset\ell_1^{\bot\bot}\subset P^\bot,
$$
and similarly $\ell_2\subset P^\bot$. Then $\ell_1+\ell_2\subset P^\bot$ and so 
$P=P^{\bot\bot}\subset(\ell_1+\ell_2)^⊥$, as desired.

*If $P,Q$ are points and $P\notin Q^\bot$, then $P+Q^\bot=\Bbb{P}$. Indeed, since $Q^\bot$ is a hyperplane, there exists a point $R\notin Q^\bot$ such that
$R+Q^\bot=\Bbb{P}$. But then $P\in R+q$ for some $q\in Q^\bot$. Since $P\ne q$, we obtain $R\in P+q$, hence 
$$
\Bbb{P}=R+Q^\bot\subset P+q+ Q^\bot=P+Q^\bot.
$$

*For any subspace $\ell$ and any two points $P,Q$ with $P\in \ell\setminus Q^\bot$, we have
$$
\ell=\ell\cap Q^\bot+P.
$$
In fact, by 3. we have $\Bbb{P}=Q^\bot+P$. Now assume $X\in \ell$, then $X\in q+P$ for some $q\in Q^\bot$. But if $X\ne P$, then $q\in X+P$, and since $P,X\in\ell$, we obtain $q\in \ell$. Hence $q\in\ell\cap Q^\bot$ and so 
$X\in q+P\subset \ell\cap Q^\bot+P$, as desired.

*Now assume that $\ell$ is closed. We want to prove that $\ell+P$ is closed. If $P\in\ell$, there is nothing to prove.  Else
$\ell^⊥\cap P^⊥\ne \ell^⊥$ (since $\ell^⊥\cap P^⊥= \ell^⊥$ implies
$\ell^⊥\subset P^⊥\ \Rightarrow P\in\ell$) and so
$$
\ell^⊥=\ell^⊥\cap P^⊥+Q
$$
for some $Q\in \ell^⊥\setminus P^⊥$. But then, by 2.,
$$
\ell=(\ell^⊥\cap P^⊥+Q)^⊥=(\ell^⊥\cap P^⊥)^⊥\cap Q^⊥.
$$
On the other hand, since $P\notin Q^⊥$, we also have 
$$
(\ell+P)^{⊥⊥}=(\ell+P)^{⊥⊥}\cap Q^⊥+P.
$$ 
From $(\ell+P)^⊥=\ell^⊥\cap P^⊥$ we deduce $(\ell+P)^{⊥⊥}=(\ell^⊥\cap P^⊥)^⊥$
and so 
$$
(\ell+P)^{⊥⊥}=(\ell^⊥\cap P^⊥)^⊥\cap Q^⊥+P=\ell+P.
$$
