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.
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 $$ (\ell+P)^{⊥⊥}=(\ell^⊥\cap P^⊥)^⊥\cap Q^⊥+P=\ell+P. $$