Actually, come to think of it, there is a very simple way of showing that every subset of is compact without using the Alexander Subspace Theorem. (That a space is Noetherian if every subset is compact is, apart from being a known fact, easy to see: Indeed, if the space contained an ascending sequence of open sets that does not stabilize, then the would form a cover of that has no finite subcover.) One simply uses the basic topological fact that a space is compact if every ultrafilter on it converges: So let be a subset of and a nonprincipal ultrafilter on (the case of principal ultrafilters is trivial). It suffices to show that for some contains every set of the form containing for some (since these sets form a subbasis). But now we simply observe that, since distinct sets of the form intersect in at most one element (since any two points in the quadrangle determine at most one line), can contain at most one such set (since it is nonprincipal and can therefore not contain two sets intersecting in finitely many elements) — say, . Therefore it must contain all sets of the form containing for each .

]]>Yes that works. So I can proceed to the next part…

]]>You‘re right, I forgot the detail that the x‘ and y‘ in the proof must be chosen collinear, with L the line through them (so there will be no and , only one L. Then the proof works: Their perps will be disjoint apart from L, and removing L will make the open sets disjoint. But we can always choose x‘ and y‘ to be collinear: after all, x but not y is collinear with x‘, and y must be collinear with a point y’ on every line containing x‘ — which x will then not be collinear with. So the proof of Hausdorffness works like this: We want to show that x and y have disjoint neighbourhoods. Take a point x‘ collinear with x but not with y and a line L containing x‘ but not x and let y‘ be the point on that line collinear with y. Then the perps of x‘ and y‘ will contain x and y, respectively, and removing L from them will make them disjoint. Does that make sense?

]]>I agree that the topology is T_1, but I don’t see Hausdorff. ]]>

Also (I should really proofread more carefully!) Noetherian means that every sequence stabilizes. I forgot the inclusions. Again, a spelling mistake, not a gap in the proof.

]]>Also, I made two slight errors in the presentation of the DeBruijn-Erdös part: Of course a point on can be collinear with two distinct collinear points not on . But then the line determined by these points must intersect and can therefore not be contained in , since was chosen disjoint from .

Also I wrote „subgraph“ at one point where I really meant „subhypergraph“.

All this is not really of crucial importance, since I have already supplied a different proof that is Noetherian. But as I said, this one has the advantage of not needing the Alexander Subbase Theorem and is also nicely geometric: points in the quadrangle become hyperplanes, and the intersection of two such hyperplanes (if it intersects ) contains a point corresponding to the line determined by them.

]]>