The 9 pseudoline counterexample means my prior definition of linear space is inadequate for constructibility. A sufficient condition for constructibility is that the space has a cospace. Then the proof of constructibility is trivial.
If "linear" space S with m boundaries has cospace C then S is constructible as hyperplanes.
Assume the conclusion for spaces with only m-1 boundaries. Construct the hyperplaces for the subspace S' of S with one boundary b removed. The cospace of S' exists because the vertex of b in the cospace of S is in a superregion that exists as a region in the cospace of S'. The cospace of S' is constructible, so choose a point in the superregion. That is a consruction of b, as required.
The cospace condition is also necessary because a constructible space has a constrctible cospace.
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