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UCR MATH 133 - Irreducibility of ordered planes

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Irreducibility of ordered planes The document http://math.ucr.edu/~res/math133/nonmetric_models.pdf states the theorem stated below; our purpose here is to give the proof together with some drawings which may be helpful for understanding the argument. Theorem. Suppose that P is plane which satisfies the standard axioms of incidence and order ( i.e., the betweenness and plane separation axioms). If Q is a flat, noncollinear subset of P, then Q = P. For our purposes here, one especially important feature of such geometrical systems is that the standard Crossbar Theorem is true in all such geometrical systems (see Theorem I I.3.5 in the course notes); proofs of this result using only the betweenness and separation axioms are is are given in pages 82 of Moïse and 116 – 117 of Greenberg. Proof. As in the related document http://math.ucr.edu/~res/math133/irreducibleplanes1.pdf (which proves the analogous result for all projective planes and all but one affine plane), we are given that three noncollinear points A, B, C lie in the flat subset Q. It follows immediately that the lines AB and BC are both contained in Q. Starting with this, we need to show that every point of P must also lie in Q. There are several steps in this process; at each step we show that Q contains more points of P than were known at the preceding one, and ultimately we find that all points of P must be in Q. If we choose D and E such that A∗B∗D and C∗B∗E hold, then by flatness we know that both D and E must lie in Q. By flatness, it follows that the lines AC, CD, DE and EA are also contained in Q, and of course this means that the segments (AC), (CD), (DE) and (EA) are also contained in Q.The next (very crucial!) step is to show that the entire interior of ∠∠∠∠ABC is contained in Q, and the drawing above suggests the argument. Given a point X in the interior of ∠∠∠∠ABC, the Crossbar Theorem implies that the ray (BX and the segment (AC) meet at some point Y. Since this point lies on AC, it must belong to Q. But we already know that B belongs to Q, and therefore the entire line BY, which contains X, must be contained in Q. Since X was an arbitrary point in the interior of ∠∠∠∠ABC, this proves the latter is contained in Q. If we switch the roles of C and E in the preceding argument, we also find that the entire interior of ∠∠∠∠ABE is contained in Q. Combining this with previously derived information, we see that all points on the same side of BC as A must belong to Q (observe that such a point is either on AB, on the same side of AB as C, or on the same side of AB as E). Finally, if we switch the roles of A and D in the preceding argument, we also find that all points on the same side of BC as D must also belong to Q. Since all points of P either lie on BC, the same side of BC as A, or the same side of BC as D, it follows that every point in P must belong to Q, so that Q =


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