Continuous projective geometry on the interval Put $P=[0,1]$. Is there a compact subset $L$ of the hyper space of $P$ such that the pair $(P, L)$ satisfies  the following axioms of projective geometry. Furthermore  the obvious maps from the configuration space of $P$ to $L$ and configuration space of $L$ to $P$ would be continuous?
Non-isomorphic projective planes on $\omega$
 A: If $(P,L)$ is an abstract projective plane, then for any point $p\in P$ and any line $\ell\in L$ not incident to $p$ there is a bijection between the set of points incident to $\ell$ and the set of lines incident to $p$. Under a reasonable definition of "topological projective plane" this bijection should be a homeomorphism.
If $P=[0,1]$, then the space of all lines incident to $0$ has a continuous surjection from the connected space $(0,1]$. It follows that for each line $\ell$ the space of all points incident to $\ell$ is connected. But one of these lines has both $0$ and $1$ in it, and the only connected subset of an interval containing both endpoints is the whole interval. Contradiction.
A: As a useful (I hope) step toward solving Ali Taghavi's problem, let me propose the following general definition of a topological projective plane:
DEFINITION   A topological projective plane is an ordered quadruple $\ \mathbf P\ :=\ (\,P\ L\ \vee \wedge\,)\ $ such that $\ P\ $ and $\ L\ $ are topological spaces, and $\ \vee:\binom P2\rightarrow L\ $ and
$\ \wedge:\binom L2\rightarrow P\ $ are continuous functions,
and $\ \mathbf P\ $ is a projective plane in the usual (non-topological, just abstract) sense, i.e. the following axioms hold:

*

*$\ \forall_{\{a\ b\ c\}\in\binom P3}\ 
           (\,a\vee b = b\vee c\ \Rightarrow\ a\vee c=a\vee b\,) $


*$\ \forall_{\{A\ B\ C\}\in\binom L3}\ 
         (\,A\wedge B = B\wedge C\ \Rightarrow\ A\wedge C = A\wedge B\,) $


*$\ \forall_{\{A\ B\ C\}\in\binom L3}\ \left( A\wedge B=A\wedge C\quad or \quad (A\wedge B)\vee(A\wedge C) = A \right) $


*$\ \forall_{\{a\ b\ c\}\in\binom P3}\ \left(a\vee b=a\vee c\quad
     or  \quad (a\vee b)\wedge(a\vee c) = a \right) $


*$\ \exists_{E\in\binom P4}\  
    \left|\left\{x\vee y: \{x\ y\}\in\binom E2\right\}\right| = \binom 42 $

REMARK   Axioms 1 and 2 can be written in the style of axioms 2 and 3 as follows:


1'. $\ \forall_{T\in\binom P3}\ 
        \left|\left\{x\vee y: \{x\ y\}\in\binom T2\right\}\right| \ =\ 1\ \text{or}\ \ 3  $


2'. $\ \forall_{t\in\binom L3}\ 
        \left|\left\{X\wedge Y: \{X\ Y\}\in\binom t2\right\}\right|\ =\ 1\ \text{or}\ 3  $

Now one can impose additional constraints, perhaps topological, on the notion of the topological projective plane to obtain more specialized (narrower) classes.

Acknowledgment    The definition here is a simplification and generalization of the definition given by Ali Taghavi from the Question above.


                **ADDITIONAL DEFINITIONS**
Now we may define the induced linear sets and pencils
$$ \forall_{A\in L}\ \ _\{A_\}\ :=\ \bigcup \vee^{-1}(A)\ =\ \bigcup
     \left\{ \{a\ b\}\in \binom P2: a\vee b = A \right\} $$
and
$$ \forall_{a\in P}\ \ ^\{p^\}\ :=\ \bigcup \wedge^{-1}(a)\ =\ \bigcup 
     \left\{ \{A\ B\}\in \binom L2: A\wedge B = a \right\} $$

The (default) topology in $\ \binom X2\ $ is induced by the canonical
map $\ X\times X\setminus\{(x\ x):x\in X\}\rightarrow\binom X2\ $ given by $\ (x\ y)\mapsto\{x\ y\},\ $ where $\ X\ $ is $\ P\ $ or $\ L$.
