One succinct statement of the projection theorem in Hilbert space is $A+A^\bot=\scr H$, where $A\in\scr C$, the set of closed subspaces of $\scr H$. (We will also denote the set of all subspaces by ${\scr S}$.) However, the proof of this statement probably requires countable choice (see my other question), so I am interested in the structure that can be obtained without assuming it. It is a consequence of this theorem that:
- $A\in{\scr C}\to A^{\bot\bot}=A$
- $A\in{\scr S}\to A^{\bot\bot}=\overline A$
- $A,B\in{\scr C},A\subseteq B^\bot\to A+B\in{\scr C}$
- $A,B\in{\scr C},A\subseteq B^\bot\to A+B=A\vee B$
In the above, $A+B=\{x+y:x\in A,y\in B\}$ is the subspace sum, and $A\vee B=(A\cup B)^{\bot\bot}$ is the subspace join. Facts that are known without the projection theorem:
- $A\in{\scr S}\to A\subseteq A^{\bot\bot}$
- $A\in{\scr S}\to A^\bot\in{\scr C}$
- $A,B\in{\scr S}\to (A\subseteq B^\bot\leftrightarrow B\subseteq A^\bot)$
There are of course many others. I'm trying to figure out exactly what picture to have of the double orthocomplement without this key theorem. Most of the theorems that use the projection theorem involve closed subspaces in some way; any non-closed subspace must have $A\subsetneq A^{\bot\bot}$ because $A^{\bot\bot}$ is closed; but without the projection theorem this might be true also for closed subspaces. Is it possible to show anything about how much larger $A^{\bot\bot}$ can be? Does $A^{\bot\bot\dots}$ ever reach a fixed point? Are there any results in the first list that can be moved to the second list?