paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0106093 | We start by proving the equivalence of REF . Any isomorphism between two graphs MATH and MATH induces a color preserving isomorphism between the two complete graphs constructed from MATH and MATH in the first step. Of course, such a color preserving isomorphism induces a color preserving isomorphism of the graphs of the polytopes constructed in the second step, which finally gives rise to an isomorphism of the graphs MATH and MATH of the two polytopes constructed in the third step. In order to prove the converse direction, let MATH and MATH be two graphs on MATH and MATH nodes (MATH), respectively, and let MATH be an isomorphism between MATH and MATH. Since MATH is MATH-regular and MATH is MATH-regular, we have MATH. If MATH, then both MATH and MATH are cycles of length MATH. Since in this case, the number of edges of MATH as well as of MATH must be MATH, MATH and MATH are isomorphic. Thus, we may assume MATH. We consider MATH and MATH colored as defined in the description of the construction. In both graphs, all MATH-cliques are node-disjoint. Each of these cliques either consists of green or of blue edges (blue cliques might arise from isolated nodes). Consider the graphs that arise from MATH and MATH by shrinking all MATH-cliques. Those nodes that come from shrinking green cliques are contained in (maximal) MATH-cliques in the shrunken graphs, while those coming from blue cliques are not (notice that for graphs without edges this statement indeed only holds for maximal MATH-cliques). This shows that MATH preserves the colors of MATH-cliques. Let MATH and MATH be the graphs that are obtained from shrinking the green cliques in MATH and MATH, respectively. Since MATH maps green cliques to green cliques, it induces an isomorphism MATH between MATH and MATH. Since the shrinking operations do not generate multiple edges, the graphs MATH and MATH inherit colorings of their edges from MATH and MATH, respectively. Because an edge of MATH or MATH is red if and only if it is not adjacent to a green edge, the isomorphism MATH preserves red edges. In the graphs MATH and MATH the only MATH-cliques are the ones formed by the blue edges. Again, these cliques are pairwise node-disjoint. Thus the isomorphism MATH between MATH and MATH induces a color preserving isomorphism between the (complete) graphs obtained by shrinking all MATH-cliques in MATH and MATH (which, again, does not produce multiple edges). This, finally, yields an isomorphism between MATH and MATH. The equivalence of REF follows from REF and CITE (see also NAME 's beautiful proof CITE) stating that two simple polytopes are isomorphic if and only if their graphs are isomorphic. For the special polytopes arising from our construction, the equivalence can, however, be alternatively deduced similarly to the proof of `` REF ." REF obviously are equivalent. Unlike the situation for simple polytopes, it is, in general, not true that two (simplicial) polytopes are isomorphic if and only their graphs are isomorphic. Nevertheless, for stacked polytopes like MATH and MATH it is true (this follows, for example, from the fact that one can reconstruct the vertex-facet incidences of a stacked MATH-polytope from its graph by iteratively removing vertices of degree MATH). Thus, finally the equivalence of REF is established. |
math/0106093 | The key observation for the proof is that every isomorphism of two simple polytopes is determined by its restriction to an arbitrary vertex and its neighborhood. Throughout the following, assume MATH. Let MATH be a simple MATH-polytope. We denote by MATH the set of neighbors of the node MATH in MATH and define MATH. The MATH-skeleton of MATH induces, for each edge MATH of MATH, a bijection MATH with MATH, MATH, and MATH being the other (than MATH) neighbor of MATH in the MATH-face spanned by MATH, MATH, and MATH. Suppose MATH is an isomorphism of the graphs MATH and MATH of two simple MATH-polytopes MATH and MATH, respectively. For each MATH we denote by MATH the restriction of MATH to MATH. For every node MATH, we have MATH and, for every edge MATH, MATH (where the composition MATH is well-defined due to REF ). Conversely, for two simple MATH-polytopes MATH and MATH with MATH vertices consider any set of maps MATH (MATH). If the maps are consistent, that is, MATH then there is a unique map MATH such that MATH is the restriction of MATH to MATH for all MATH. We claim that, if MATH satisfies REF for all MATH, then MATH is an isomorphism of MATH and MATH (and thus a combinatorial isomorphism between MATH and MATH). To see this, it suffices to show that MATH is surjective, since both MATH and MATH have the same number MATH of nodes. Suppose that MATH is not surjective. Since MATH is connected, there is a node MATH of MATH which is not contained in the image of MATH and which has a neighbor MATH that is the image MATH of some node MATH of MATH. However, this contradicts REF . Thus, we may check two simple MATH-polytopes MATH and MATH both with MATH vertices, given by their vertex-facet incidences, for combinatorial isomorphism in the following way. First, we compute the graphs MATH and MATH, as well as the bijections MATH REF and MATH (MATH). Furthermore, a node MATH is fixed together with a spanning tree MATH of MATH, rooted at MATH. Then, for each node MATH and for each bijection MATH with MATH, we perform the following steps: CASE: Compute MATH (MATH) ``along MATH" by means of REF ; if, for some MATH, REF is not satisfied, then continue with the next bijection MATH. CASE: If REF is not satisfied, then continue with the next bijection MATH. CASE: Construct the map MATH (as above); STOP. Note that, when MATH is computed in REF from the parent MATH of MATH in MATH, REF is satisfied for MATH (thus, the composition in REF is well-defined). It follows from the discussion above that the two polytopes are isomorphic if and only if the algorithm stops in REF ; in this case, the constructed MATH is an isomorphism between MATH and MATH. Computing the graphs MATH and MATH, as well as the bijections MATH REF and MATH (MATH), can be performed in MATH time, if we perform the following preprocessing in advance. From the vertex-facet incidences compute, for each vertex, a sorted list of indices of the facets containing this vertex. This can be performed in MATH steps (note that simple polytopes never have more facets than vertices). Compute a similar incidence list for each facet. Moreover, using this data structure, none of the four steps in the for-loop needs more than MATH time (the critical part being REF ). Since the body of the for-loop is not executed more than MATH times, this yields a MATH time algorithm. |
math/0106093 | The core of the proof is a result of CITE implying that every combinatorial isomorphism between the barycentric subdivisions of two polytopes MATH and MATH induces a combinatorial isomorphism between MATH and MATH or its dual polytope MATH, where the barycentric subdivision of a MATH-polytope MATH may be defined as any MATH-polytope MATH, whose vertices correspond to the non-trivial faces REF of MATH, and whose facets correspond to the maximal chains in the face lattice of MATH. Such a polytope MATH can be constructed from MATH by performing stellar subdivisions of all MATH-faces for MATH (see REF ). The dual MATH is a simple MATH-polytope whose vertices correspond to the maximal chains in the face lattice of MATH, and whose facets correspond to the (non-empty) faces of MATH. In the graph of MATH, two nodes are adjacent if and only if the corresponding two maximal chains of MATH arise from each other by exchanging the two MATH-dimensional faces (for some unique MATH). This defines a labeling of the edges of MATH by MATH, where for each node the MATH incident edges receive pairwise different labels. Thus, NAME 's result implies that two polytopes MATH and MATH are combinatorially isomorphic if and only if there is a combinatorial isomorphism between MATH and MATH that induces a label preserving isomorphism of MATH and MATH. Let MATH and MATH be two MATH-polytopes, given by their vertex-facet incidences, which are to be checked for combinatorial isomorphism. We first compute the NAME diagram of the face lattice of each of the two polytopes (with nodes labeled by the dimensions of the respective faces). From the NAME diagrams we then enumerate the maximal chains of the face lattices of MATH and MATH, the vertex-facet incidences of MATH and MATH, as well as the graphs MATH and MATH together with their edge labelings. Once these data are available, we can check MATH and MATH for combinatorial isomorphism as in the proof of REF . However, since we only allow isomorphisms that respect the edge labeling of MATH and MATH, for each potential image MATH of the fixed vertex MATH of MATH there is only one possible bijection between MATH (with the notation adapted from the proof of REF ). Consequently, the running time of checking MATH and MATH for (label preserving) combinatorial isomorphism can be estimated by MATH, where MATH is the number of maximal chains in the face lattice of MATH or MATH. We may assume that MATH is the same number for both MATH and MATH, since otherwise we know already that MATH is not isomorphic to MATH; the same holds for the numbers MATH, MATH, MATH, and MATH arising below. Computing the NAME diagrams of MATH and MATH can be performed in time MATH (see CITE), where MATH, MATH, MATH, and MATH are the number of faces, the number of vertex-facet incidences, the number of vertices, and the number of facets of MATH or MATH, respectively. Enumerating all maximal chains in the face lattices of MATH and MATH, can be done in time proportional to MATH. Computing the vertex-facet incidences of MATH and MATH takes no more than MATH steps. The edge labeled graphs MATH and MATH as well as the maps MATH can be obtained (see the proof of REF ) in time MATH (note that MATH can easily be obtained from the edge labeling). By the upper bound theorem for convex polytopes (see, for example, CITE) applied to the barycentric subdivisions, we have MATH, where, again by the upper bound theorem, we can estimate MATH. In total, the running time of the sketched algorithm thus can be bounded by MATH. |
math/0106094 | We may assume that each pro-object MATH is cofinite directed with index set MATH. Choose MATH to be an arbitrary cofinite directed set with cardinality greater than or equal to the cardinalities of every MATH; this will be the index set for the level representation MATH. Also choose arbitrary set surjections MATH. For every MATH, we build a new pro-object MATH by constructing a cofinal function MATH and letting MATH equal MATH. Fix an element MATH of MATH. Assume that the function MATH has already been constructed on all indices MATH for which there exists a map MATH in MATH. Then we may proceed by induction because MATH is cofinite. We may define MATH inductively since MATH is cofinite. Let MATH be an index in MATH, and suppose that MATH has already been defined for MATH. We choose MATH satisfying the following properties. This is possible because there are only finitely many conditions. First, choose MATH sufficiently large so that MATH for all MATH. This guarantees that MATH is a pro-object. Second, choose MATH large enough so that MATH. This guarantees that MATH is cofinal so that the natural map MATH is an isomorphism. Third, choose MATH large enough so that for all maps MATH in MATH, there are maps MATH representing MATH such that the diagrams MATH commute for all MATH. This guarantees that MATH is a level representation. Finally, choose MATH large enough so that for all pairs of arrows MATH and MATH in MATH, the diagram MATH commutes. This guarantees that MATH is a level representation. Note that the isomorphisms MATH are given by representatives MATH that are identity maps. It follows that the diagram MATH commutes for every map MATH in MATH because both compositions are given by representatives MATH . |
math/0106094 | This follows immediately from REF . |
math/0106094 | Because each MATH is isomorphic to MATH, it suffices to show that MATH is isomorphic to MATH. By direct calculation with the definition of morphism sets for MATH, MATH satisfies the required universal property. |
math/0106094 | First we show that the class of pro-objects that are essentially of type MATH is closed under cofiltered limits. Let MATH be a cofiltered diagram of pro-objects, each of which is essentially of type MATH. We may assume that each MATH is actually of type MATH. We use the method of CITE to replace each MATH by a cofinite directed pro-object that is still of type MATH. Then we use this same method to replace the diagram MATH by a cofinite directed diagram with the same limit such that each MATH is still a cofinite directed pro-object of type MATH. Now we use the construction of REF . Since each MATH belongs to MATH, each MATH belongs to MATH (see the proof of REF ). Since each MATH belongs to MATH, the pro-object MATH is of type MATH. Therefore, MATH is essentially of type MATH. Now we must show that every pro-object that is essentially of type MATH is isomorphic to a cofiltered limit of objects in MATH. It suffices to show that every pro-object of type MATH is a cofiltered limit of objects in MATH. Let MATH be a pro-object such that each MATH belongs to MATH. By direct calculation of morphism sets, MATH is isomorphic to MATH. Also, each MATH belongs to MATH by assumption. |
math/0106094 | Apply REF to the category MATH. We use that the categories MATH and MATH are equivalent. |
math/0106094 | Let MATH be essentially of type MATH, and let MATH be another pro-object with maps MATH and MATH such that the composition MATH is the identity. Consider the countable tower MATH . Since MATH is the identity, the limit of this tower is isomorphic to MATH. On the other hand, the limit is also isomorphic to the limit of the tower MATH . By REF , this limit is essentially of type MATH. |
math/0106094 | Apply REF to the category MATH and use that MATH and MATH are equivalent. |
math/0106094 | Let MATH be a cofiltered index category, and let MATH be a finite index category. Suppose given a functor MATH . In order to compute limits with respect to MATH, we may replace MATH with a cofinite cofiltered index category as described at the beginning of REF. In order to compute colimits with respect to MATH, we may replace MATH by a finite loopless category; this is the usual method for rewriting any finite colimit in terms of finite coproducts and coequalizers. Therefore, we may assume that both MATH and MATH are cofinite. It follows that MATH is cofinite, so we may assume that MATH is a level representation. Thus, we have a functor MATH . Recall that finite colimits of pro-objects may be computed levelwise. By REF and direct computation, both MATH and MATH are isomorphic to the pro-object MATH . |
math/0106094 | Limits are always left exact. NAME limits are right exact by REF . |
math/0106094 | First consider a constant pro-object MATH. Let MATH be an arbitrary cofiltered system of pro-objects indexed by MATH. As described at the beginning of REF, we may assume that MATH is cofinite. Thus, we may take MATH to be a level representation. It follows by REF and direct calculation that both MATH and MATH are equal to MATH. This finishes one direction of the theorem. Now suppose that MATH is a cocompact pro-object. Note that MATH is isomorphic to MATH, so the map MATH is an isomorphism. Let MATH be any map such that MATH is equal to the identity on MATH, and let MATH be the natural map. By definition of MATH, the composition MATH is the identity on MATH. In order to show that MATH and MATH are isomorphic, it suffices to show that the composition MATH is the identity on MATH. Define a map MATH by the formula MATH. Because MATH is the identity, it is easy to check that MATH is the inverse of MATH. Since MATH and MATH both equal MATH, it follows that MATH is the identity map. |
math/0106094 | Let MATH and MATH be any two elements of MATH. We construct a common refinement MATH by induction on MATH. This is possible since MATH is cofinite. Suppose that MATH has already been determined for MATH such that MATH and MATH. Choose MATH such that MATH, MATH, and MATH for all MATH. This is possible because there are only finitely many conditions on MATH. |
math/0106094 | Given an element MATH of MATH, we must find an element MATH of MATH such that MATH. Suppose that MATH has already been chosen for MATH such that MATH. Choose MATH such that MATH and MATH for all MATH. This is possible since MATH is cofinite. |
math/0106094 | The projection functor MATH is cofinal. By REF , the functor MATH is a composition of two cofinal functors, so it is also cofinal. |
math/0106094 | Let MATH be an arbitrary pro-object. Then direct computation with the definition of morphism sets in MATH shows that MATH . This uses REF and the surprising fact that the canonical map MATH is an isomorphism, which uses REF . |
math/0106095 | Since a MATH-degree rotation about the MATH-axis maps MATH to MATH and leaves the helix invariant, the bitangent sphere must be centered on the MATH-axis. Thus, MATH can be described by the equation MATH for some constants MATH and MATH. Let MATH denote the intersection curve of MATH and the cylinder MATH. Every intersection point between MATH and the helix must lie on MATH. If we project the helix and the intersection curve to the MATH-plane, we obtain the sinusoid MATH and a portion of the parabola MATH. These two curves meet tangentially at the points MATH and MATH. The mean value theorem implies that MATH at most four times in the range MATH. (Otherwise, the curves MATH and MATH would intersect more than twice in that range.) Since the curves meet with even multiplicity at two points, those are the only intersection points in the range MATH. Since MATH is concave, we have MATH, so there are no intersections with MATH. Thus, the curves meet only at their two points of tangency. |
math/0106095 | Call a tetrahedron with vertices MATH local if MATH and MATH. Let MATH be the sphere passing through the vertices of an arbitrary local tetrahedron. Analysis similar to the proof of REF implies that the only portions of the helix that lie inside MATH are the segments between MATH and MATH and between MATH and MATH. Thus, all other points in MATH lie outside MATH, so the four points form a NAME simplex. The local NAME simplices exactly fill the convex hull of MATH, and therefore comprise the entire NAME triangulation. Specifically, the only triangles that are facets of exactly one local tetrahedron have vertices MATH or MATH for some integer MATH. Thus, a tetrahedron is NAME if and only if it is local. |
math/0106095 | Consider the sphere MATH passing through MATH arbitrary points MATH and tangent to the generalized helix at those points, where MATH. Any point MATH that lies on MATH satisfies the following matrix equation: MATH . To bound the number of zeros of MATH, consider its second derivative MATH . MATH is an affine combination of the functions MATH, so it can be rewritten as as a polynomial of degree at most MATH in the variable MATH. Thus, MATH has at most MATH zeros in the range MATH. (This is essentially the argument used by NAME to show that Petrie polytopes are neighborly CITE.) Since MATH are roots of MATH of multiplicity two, they are the only roots in the range MATH; otherwise, by the mean value theorem, MATH would have more than MATH roots in the range MATH, which we have just shown to be impossible. Thus, the points MATH lie on a sphere that excludes every other point in MATH and so have mutually neighboring NAME regions. |
math/0106095 | Fix an integer MATH, and for notational convenience, let MATH. Since MATH is preserved under a rigid motion mapping each point MATH to its successor MATH, the NAME regions of MATH are congruent. Call a full-dimensional simplex with vertices in MATH local if all its vertices consist of MATH adjacent pairs within a single turn of the generalized helix, that is, if its vertices are MATH for some integers MATH with MATH and MATH for all MATH. Analysis similar to REF implies that every local simplex is NAME. The convex hull of MATH, which we will call the Petrie cylinder, is the product of a MATH-dimensional Petrie polytope with MATH vertices and a line orthogonal to that polytope's hyperplane. By NAME 's evenness condition CITE, the facets of the Petrie polytope are formed by all sets of MATH adjacent pairs of points on the trigonometric moment curve. The faces of the Petrie cylinder are cylinders over the faces of the Petrie polytope. Call a facet of a local simplex that is not shared by another local simplex a boundary simplex. We easily observe that the boundary simplices are exactly the MATH-simplices whose ordered sequence of vertices has one of the following two forms: MATH . The following sequence of boundary simplices exactly covers one facet of the Petrie cylinder. MATH . Every facet of the Petrie cylinder is covered in this manner, and every boundary simplex lies on some facet of the Petrie cylinder. Thus, the union of the boundary facets is the boundary of the Petrie cylinder, so the local NAME simplices completely fill the Petrie cylinder and therefore comprise the entire NAME triangulation. It easily follows that each NAME region of MATH is a convex polyhedron with MATH facets, and that any MATH consecutive NAME regions form a MATH-neighborly family. As we already observed, these polyhedra are congruent. |
math/0106100 | CASE: MATH. We only need to show that MATH is entailed by MATH. Suppose MATH. So MATH or MATH, by MATH. But MATH and MATH, by MATH. CASE: MATH CASE: MATH. If MATH, there is a MATH such that MATH and MATH by MATH. If MATH, then MATH because MATH, by MATH. So, there is a MATH such that MATH and MATH. So MATH and, by MATH, MATH. But then MATH by MATH. CASE: MATH. If MATH and MATH, then MATH, by MATH, contra MATH. CASE: MATH, MATH, and MATH follow from the corresponding principles for MATH and MATH. CASE: MATH, MATH, MATH, and MATH are logical consequences of MATH. |
math/0106100 | In both cases, the finite cardinalities determine a quasi-linear ordering of the sets in which any set is higher than any of its proper subsets. |
math/0106100 | We shall prove REF by induction on MATH: MATH . Now, suppose that REF is true for all MATH: MATH . |
math/0106100 | MATH . |
math/0106100 | MATH . |
math/0106100 | The proofs are elementary. |
math/0106100 | MATH. By compactness, then, there is a MATH such that MATH . So MATH is true in any atomic boolean algebra with more than MATH atoms. |
math/0106100 | If MATH, then MATH is true is some atomic boolean algebra, MATH. If MATH is finite, then MATH is isomorphic to some standard finite interpration MATH of MATH. But, then MATH is true in MATH, so MATH is not true in MATH and MATH. If MATH is infinite, then MATH is consistent with MATH. But MATH is complete, so MATH. By REF, MATH is true in some finite model MATH of MATH. Hence, MATH is true in some standard finite interpretation of MATH and, again, MATH. |
math/0106100 | If MATH, MATH. The model, MATH defined above satisfies MATH but not MATH, for any MATH finite sets have a finite union and any two cofinite sets overlap. So, MATH, by REF. |
math/0106100 | CASE: Suppose MATH and MATH. Then MATH. CASE: MATH itself is the union of MATH-congruence classes. If MATH is the union of MATH of these classes, then MATH is the union of the remaining MATH-congruence classes. |
math/0106100 | CASE: Let MATH and let MATH. CASE: Suppose MATH. Then MATH and MATH, so MATH and MATH are finite. |
math/0106100 | Since MATH, MATH. But MATH is finite, so MATH is finite. Since MATH, MATH. But MATH is finite, so MATH is finite. |
math/0106100 | CASE: MATH is near MATH, since MATH, which is finite. CASE: If MATH is near MATH, then MATH is near MATH. Immediate. CASE: Suppose MATH is near MATH and MATH is near MATH. Note that MATH . But MATH is finite because MATH is finite and MATH is finite because MATH, which is finite. So the union, MATH, is finite. Similarly, MATH where MATH and MATH. So MATH is also finite. Hence, MATH is near MATH. |
math/0106100 | CASE: Since MATH is near MATH, MATH and MATH are finite. But MATH CASE: By REF , MATH is near MATH , which is near MATH. So MATH is near MATH, by transitivity. CASE: MATH and MATH. So if MATH and MATH are near each other, so are their complements. |
math/0106100 | Using REF : CASE: MATH, since MATH and if MATH, then MATH. CASE: If MATH, then MATH . CASE: MATH . |
math/0106100 | CASE: Let MATH be the least number such that MATH . So each is a disjoint union of MATH-congruence classes: MATH . Similarly, if MATH, then MATH is infinite. CASE: If MATH were near two quasi-congruence classes, the two would have to be near each other, since NEAR is transitive. But this is impossible by REF . |
math/0106100 | CASE: If MATH for some MATH, MATH, then, for some MATH, MATH . So MATH. If MATH, there is a MATH such that MATH. Let MATH be the greatest MATH such that MATH and let MATH. MATH CASE: By REF , each MATH-congruence class is a disjoint union of MATH-congruence classes. CASE: Suppose MATH and MATH. Then, by REF , both MATH and MATH are in MATH. |
math/0106100 | By REF, MATH is an atomic boolean algebra; so MATH. The MATH-relation of MATH is induced from the linear ordering of sizes; so it is a quasi-linear ordering and MATH, MATH, and MATH are satisfied. As for the remaining axioms: CASE: MATH: Suppose MATH. If MATH then MATH and MATH and MATH, where at least one of these inclusions is proper. So MATH. But if MATH, then MATH, so MATH. In either case, MATH, so MATH. CASE: MATH: Suppose MATH. So MATH, MATH, and either MATH or MATH. We want to find some MATH such that MATH. If MATH, then MATH must be infinite; so, MATH can be obtained by removing MATH atoms from MATH. If MATH, then MATH; so, again, MATH can be obtained by removing MATH atoms from MATH. If MATH, then let MATH be the union of MATH-congruence classes contained in MATH. So MATH is infinite and MATH is finite. Let MATH. So MATH where MATH. Finally, let MATH and MATH, where MATH with MATH. If MATH, then MATH is finite. If MATH, then MATH is infinite, so there is no problem. If MATH, then MATH is finite, but has more members than MATH, since MATH. So, let MATH be any proper subset of MATH with MATH members. CASE: MATH: It is enough to show that if MATH and MATH are disjoint, then MATH. We need the following three facts: CASE: MATH. (See REF .) CASE: MATH. If MATH but MATH, then MATH or MATH; any element of MATH is also in MATH unless it is in MATH; any element of MATH is also in MATH unless it is in MATH. CASE: MATH. Note that if MATH, MATH, and MATH, then MATH; otherwise, MATH and MATH have an infinite intersection. MATH . From REF we obtain (iia): MATH and from REF we obtain (iiia): MATH . But MATH and MATH are disjoint, since MATH and MATH are disjoint. So: MATH . Since MATH and MATH are contained, respectively, in MATH and MATH, which are disjoint, they are also disjoint. So: MATH . So, MATH . Since MATH, by REF , we know that MATH. CASE: MATH: MATH (Cancellation is valid for sizes because it is valid for rationals and integers.) Conversely, MATH . |
math/0106100 | CASE: MATH. Hence, if MATH disjoint sets of the same size exhaust MATH , they must each have size MATH. But if MATH, then MATH, for integral MATH and MATH. So MATH and MATH. CASE: For each MATH, MATH, let MATH. So MATH. If MATH, then MATH and MATH since MATH. Letting MATH, group the MATH sets MATH into MATH collections with MATH members in each: MATH . Letting MATH for MATH, we have MATH and MATH. Furthermore, MATH. |
math/0106100 | Based on REF : CASE: If MATH, then MATH CASE: MATH since it satisfies MATH for each MATH. But, if MATH, then MATH. CASE: Immediate from REF , since only finitely many MATH divide MATH. |
math/0106100 | Every set in every standard finite interpretation is a finite set, and all finite sets are roughly divisible by every MATH. |
math/0106100 | CASE: By induction on MATH: if MATH, then MATH, so MATH. If MATH, MATH, and MATH, then MATH can be partitioned into MATH sets of the same size and MATH atoms, where MATH. Each non-atomic set in the partition can be further partitioned into MATH sets of the same size and atoms, where MATH. Thus, we have partitioned MATH into MATH sets of the same size and MATH atoms. But MATH and, since MATH and MATH, MATH. Hence, MATH. CASE: Obvious. CASE: Immediate from REF . |
math/0106100 | If MATH, then by compactness there is a finite set MATH such that MATH. But then MATH for every MATH, by REF . But this contradicts REF , which says that MATH entails only finitely many MATH sentences. |
math/0106100 | CASE: CITE CASE: Suppose that MATH, MATH. Then MATH is consistent. MATH has a consistent, complete extension, MATH, by NAME 's lemma. Since MATH, MATH is also a consistent, complete extension of MATH. But MATH is not consistent with MATH. |
math/0106100 | CASE: Suppose MATH is a finite model of MATH with MATH atoms. Since MATH is complete, MATH. So MATH and has no infinite models. CASE: MATH for every MATH, so MATH has no finite models. |
math/0106100 | CASE: MATH and, by REF , MATH is categorical in every finite power. CASE: The model MATH of MATH is a standard finite interpretation. So MATH. |
math/0106100 | CASE: MATH . Let MATH, Since MATH is consistent, there is some MATH such that MATH. So MATH is a solution of MATH and, hence, MATH is congruous by REF. MATH . If MATH is congruous, MATH has a solution, MATH. So MATH CASE: MATH . For every finite restriction, MATH, of MATH, MATH is consistent. By REF , each such MATH is congruous. Hence, MATH is congruous by REF. MATH . Every finite restriction, MATH, of MATH is congruous. So MATH is consistent, by REF . By compactness, then, MATH is consistent. CASE: MATH . If MATH is consistent, so is MATH. So, by REF , MATH is congruous. MATH . By compactness, it is sufficient to show that every finite subtheory, MATH, of MATH is consistent. But, if MATH is such a theory, then MATH for some MATH and some finite restriction, MATH, of MATH. Since MATH is congruous, MATH is as well, by REF. So MATH has arbitrarily large solutions and MATH has finite models large enough to satisfy MATH. Hence, MATH is consistent. |
math/0106100 | Suppose MATH, MATH, and MATH. So, MATH can be partitioned into MATH sets of the same size MATH and MATH atoms MATH . For MATH, let MATH . Since MATH, MATH, for MATH. Furthermore, MATH . So, MATH. |
math/0106100 | Suppose MATH. Then: MATH where the MATH's and MATH's are atoms and the MATH's (MATH's) are disjoint sets of the same size in MATH. Let MATH . We claim that MATH. For, if MATH, then MATH and if MATH, then MATH; neither is possible since MATH. So MATH for MATH. Since MATH, there is a proper subset MATH of MATH which is the same size at MATH. MATH . But MATH, since each is the disjoint union of MATH atoms. Thus MATH, by MATH, so MATH. But MATH, since for each component of MATH, there is a component of MATH of the same size. So MATH, by MATH. But MATH must be larger than MATH, for MATH is the union of fewer than MATH atoms while MATH is the union of MATH non-empty sets. So, the original supposition entails a contradiction. |
math/0106100 | CASE: MATH . Supposing that MATH is incongruous, there exist MATH, MATH, and MATH such that MATH and MATH. We will show that MATH from which it follows that MATH is inconsistent. Let MATH and MATH be such that MATH . By REF , we have MATH since MATH and MATH. Since MATH, MATH. So REF yields MATH . From REF , and REF we may conclude REF . MATH . Follows from REF since MATH. CASE: See proof of REF. CASE: See proof of REF. |
math/0106100 | Immediate from REFEFc. |
math/0106100 | MATH fails to axiomatize MATH for the same reason that MATH fails to axiomatize MATH: the lack of divisibility principles. |
math/0106100 | CASE: CITE CASE: If MATH has no function symbols, then any finitely generated structure over MATH is finite. So, suppose MATH contains MATH (from MATH) and MATH (from MATH). Let MATH be the diagram of MATH and obtain MATH from MATH by substituting the variable MATH for each constant MATH and the variable MATH for each constant MATH. Finally, let MATH be MATH . So, MATH is a primitive formula. MATH, so MATH does as well, by REF in the Appendix. So, to obtain the desired isomorphism, map the MATH's into themselves and map the MATH's into a sequence of elements of MATH which can stand in for the existentially quantified variables of MATH. |
math/0106100 | Suppose MATH where MATH is a conjunction of atomic formulae and negations of atomic formulae. Then MATH . |
math/0106100 | By REF in the Appendix, it is enough to show that given models MATH and MATH of MATH, where MATH and a primitive formula, MATH: MATH . Since MATH is uniformly MATH-reducible to MATH, there are models of MATH, MATH, where MATH and MATH . |
math/0106100 | CASE: Every abelian semigroup with cancellation can be isomorphically embedded in an abelian group CITE. It is clear from the construction in CITE that if the semigroup is ordered, the abelian group in which it is embedded may also be ordered and that the elements of the semigroup will be the positive elements of the group. Moreover, the (rough) divisibility of the elements in the semigroup will be carried over to the group. Consequently, the theory of MATH -semigroups is uniformly reducible to the theory of MATH groups by the translation: MATH . Since the latter is model complete, so is the former, by REF. CASE: First, consider the theory of MATH -semigroups in the language which contains, besides the constant symbols in the original theory, an individual constant, MATH. The theory of MATH -semigroups is model-complete in this language, by REF. We claim that the theory of MATH-groups modulo MATH is uniformly reducible to this new theory by the following translation: MATH (The construction: given a model of MATH, stack up MATH many copies of the model, assigning interpretations in the obvious way. The result is a MATH -semigroup and the original model is isomorphic to the first copy of itself.) |
math/0106100 | The proofs are elementary. |
math/0106100 | Omitted. |
math/0106100 | CASE: MATH . Obvious. CASE: MATH CASE: MATH CASE: MATH . |
math/0106100 | CASE: Apply REF . CASE: Immediate from REFEF . |
math/0106100 | Proof of REF Given a molecule, MATH, let MATH be the atoms of MATH contained in MATH. For each MATH, let MATH be some member of MATH . These elements will be elements of MATH, since MATH yields sizes in MATH whose members are in MATH. These elements have the right sizes, as we will show below, but they are in the wrong places. We have no guarantee that they are contained in the molecule MATH. So we still need to show that there are disjoint elements of MATH, MATH whose union is MATH and whose sizes are the same as those of MATH, respectively. Well, MATH since each MATH and MATH. So, the existence of MATH, as above, is guaranteed because MATH. Now, let MATH for each MATH in the molecule MATH. Repeating this procedure for each molecule yields a value of MATH for each atom of MATH. Finally, if MATH is non-atomic, then MATH where each MATH is atomic. So let MATH satisfies the requirements of REF : Boolean relations are preserved by MATH because the function is determined by its values on the atoms of MATH; MATH maps elements of MATH into themselves because the set of atoms contained in each of these molecules is mapped into a disjoint collection of elements of MATH whose union is the same molecule; so, we need only show that MATH preserves size relations. To do so, we invoke REF , below: MATH . |
math/0106100 | Suppose MATH where each MATH is an atom of MATH. MATH . |
math/0106100 | The proof can be obtained from the proof of REF by substituting: MATH . |
math/0106100 | It will be clearer, and easier, to establish this by example rather than by formal proof. Letting MATH and MATH, we want to show that REF-congruence classes have the same size relations in MATH as they do in MATH, for any MATH. The other elements of MATH will then fall into place, since size relations are determined by the representations of each set, MATH, as MATH, MATH, and MATH. Suppose that MATH, so that MATH . Since MATH is congruous, MATH. If MATH, then all of REF-congruence classes are common. If MATH, then MATH and MATH are the only charmed REF-congruence classes. If MATH, then all of REF-congruence classes are charmed except for MATH and MATH. In any case, MATH will include the same number of charmed REF-congruence classes as MATH, so MATH . Suppose, however that MATH, so that MATH . Here, MATH, since MATH is congruous. In any case, MATH contains exactly one more charmed congruence class than MATH, so MATH . So it goes in general. |
math/0106100 | CASE: MATH is a universal-existential theory; so it is preserved under unions of chains. (See Appendix, REF ). CASE: Each MATH-congruence class can be partitioned into MATH-congruence classes. CASE: MATH, by definition. Since MATH is an existential sentence, it is preserved under extensions. CASE: Immediate from REF , and REF . |
math/0106100 | CASE: MATH says that MATH has a MATH-partition. CASE: All common MATH-factors are the same size and satisfy the same MATH predicate. So, suppose that each common MATH-factor has MATH charmed MATH-factors and MATH common MATH-factors. By REF , MATH has MATH charmed MATH-factors and MATH common MATH-factors. Partitioning each of the MATH-factors into MATH subsets of roughly the same size yields a MATH-partition of MATH; the charmed MATH-factors of the MATH-factors are the charmed MATH factors of MATH and the common MATH-factors of the MATH-factors are the common MATH factors of MATH. Each of the common MATH factors has MATH charmed MATH-factors and each of the charmed MATH-factors has MATH charmed MATH-factors. In all, there are MATH charmed MATH-factors among the MATH-factors of MATH. But, by REF again, there are MATH charmed MATH-factors of MATH. So MATH CASE: Immediate from REF , since every charmed MATH-factor is one atom larger than each common MATH-factor. |
math/0106100 | We demonstrate REF by induction on MATH.: If MATH, then MATH, MATH, and MATH is the basis of MATH. So REF holds because any set is a MATH -partition of itself. Assume that REF holds for MATH. Then REF also holds for MATH: each node set of depth MATH has MATH immediate descendants; so, there are MATH node sets of depth MATH. Furthermore, MATH of the MATH-factors are charmed and MATH are common. By REF , each charmed MATH-factor has a MATH-partition and each common MATH-factor has a MATH-partition, where MATH (substituting MATH for MATH and MATH for MATH). So, there are MATH . In all, then , the number of charmed MATH-factors is: MATH . So, the MATH factors of the MATH-factors of the basis form a MATH-partition of the basis. That is to say, REF holds for MATH. |
math/0106100 | CASE: MATH is generated from node sets and node atoms via the boolean operations, each of which preserves MATH-nearness to quasi-nodal sets. MATH) MATH must contain finite unions of node sets as well as MATH-finite sets; so it must also contain sets obtained by adding or removing MATH-finite sets from quasi-nodal sets. CASE: The proof parallels that of REF exactly. CASE: Let MATH be the quasi-nodal set which is MATH-near MATH (see REF; Let MATH; and, let MATH. |
math/0106100 | There is at most one such MATH, by REF . To show that there is such a MATH, suppose first that MATH. Then there is some MATH such that MATH where each MATH is an (MATH)-congruence class and hence a node set in MATH. So MATH . Now, let MATH . So MATH and: MATH . If MATH is not a quasi-congruence class, then MATH where MATH is a quasi-congruence class. Let MATH be the element of MATH corresponding to MATH, as described above, and let MATH . |
math/0106100 | CASE: Immediate from the proof of REF. CASE: Suppose MATH. Then MATH . But every integer is MATH for some MATH, so MATH. CASE: Obvious. |
math/0106100 | CASE: MATH CASE: As in REF , MATH can be shown to preserve MATH, I, unions, intersections, relative complements, and proper subsets. CASE: Let MATH be the number of charmed node sets of level MATH contained in MATH. And let MATH, below, be the least MATH such that MATH and MATH are both unions of node sets of depth MATH. MATH CASE: MATH CASE: MATH . |
math/0106100 | MATH is consistent because MATH is congruous, by REF. Since MATH is model complete, by REF, and has a prime model, by REF, the prime model test (Appendix, REF ) applies. So MATH is complete. |
math/0106100 | For each MATH, MATH for exactly one MATH, MATH: Since MATH is complete, MATH or MATH for each such MATH. But if MATH, then MATH is inconsistent, since MATH and MATH. Hence MATH for at least one one MATH, MATH. But suppose MATH and MATH where MATH. Again, MATH would be inconsistent, for MATH (see REF ). So, let MATH iff MATH. Then MATH and, since MATH is complete, MATH. |
math/0106100 | CASE: Follows from REFEF. CASE: Follows from REFEFb, given that MATH. |
math/0106100 | CASE: MATH iff MATH. But MATH is a finite model. CASE: To determine whether MATH, alternate between generating theorems of MATH and testing whether MATH. |
math/0106100 | CASE: There are MATH remainder functions whose domain is the set of prime numbers. Each such function is congruous, so each corresponds to a consistent extension of MATH. By NAME 's lemma, each of these extensions has a consistent and complete extension. CASE: MATH . If MATH is decidable, then MATH is recursively enumerable. But MATH is complete, so it is decidable. MATH . To calculate MATH, determine which MATH sentence is in MATH. |
math/0106100 | If MATH is true only in infinite models of MATH, then MATH is true in all finite models of MATH, so MATH. But MATH, so MATH is inconsistent. |
math/0106100 | Suppose MATH. So MATH and, by compactness, MATH for some finite set of MATH principles, MATH. Let MATH. Let MATH be a model with domain MATH, in which size relations are determined in accordance with the size function, MATH, defined in REFEF. We claim the following without proof: MATH . By REF , MATH, so MATH. But by REF , MATH. Hence MATH. |
math/0106100 | CASE: MATH has at least MATH members less than or equal to MATH for any given MATH. MATH has at most MATH such members. If MATH, then MATH CASE: Similar to REF . CASE: If MATH, both MATH and MATH have exactly MATH members less than MATH. For MATH, MATH will have more members in MATH than in MATH. |
math/0106100 | CASE: By MATH, it is sufficient to show that MATH. If it were, then, by MATH, there is a MATH such that MATH and MATH. But any proper subset of MATH is outpaced by, and hence smaller than MATH. Let MATH be the least odd number not in MATH. Then MATH leads MATH at MATH and MATH never catches up. So there is no MATH such that MATH. CASE: Similar to that of REF . CASE: Note that MATH and that MATH REF . MATH . Let MATH, MATH, MATH and apply REF . |
math/0106100 | The argument for REFEF applies here since the only facts about MATH and MATH used hold by virtue of these sets forming an alternating pair. |
math/0106100 | CASE: MATH,MATH, MATH and MATH. CASE: If MATH for some MATH, let MATH be the least such MATH; otherwise, let MATH. CASE: For MATH, MATH is an alternating pair. So either MATH or MATH by REF. But MATH by the selection of MATH, so MATH. REF follows by MATH. CASE: Immediate from REF since MATH is an alternating pair and MATH. CASE: MATH if MATH. So MATH. Hence MATH. So MATH. |
math/0106100 | CASE: By induction on the structure of MATH: CASE: If MATH is a constant, MATH for some MATH. So MATH by REF. CASE: If MATH, MATH . The proofs for intersections and relative complements are similar. CASE: By induction on the structure of MATH: CASE: If MATH, then MATH iff MATH . If MATH, then there is a MATH but not MATH. So if MATH, MATH. Hence, MATH is cofinite and, by REF, in MATH. Conversely, if MATH, then it is infinite. So, there cannot be a MATH in MATH but not in MATH; otherwise MATH would not be included in MATH for any MATH greater than MATH. So MATH. But, clearly MATH, so MATH. CASE: If MATH, then MATH since if MATH and MATH, then MATH. CASE: MATH iff MATH. Immediate from REF. CASE: If MATH is non-atomic, then, since MATH is an ultrafilter: MATH CASE: MATH CASE: Suppose MATH. Construct MATH, a subset of MATH, for which MATH as follows: MATH . Then MATH since each MATH draws its new members from MATH. Claim: If MATH, then MATH. Hence: MATH since they are the same size over some set which contains MATH and is, thus, if MATH. CASE: Immediate from REF : MATH is equivalent to a set of universal sentences, together with MATH and MATH. (MATH because it contains all singletons of natural numbers.) CASE: For MATH, MATH an infinite subset of MATH , let MATH. The MATH sets, MATH, partition MATH. Furthermore, these form an alternating MATH-tuple, in the manner of the congruence classes modulo MATH (see REF ). As in REFEF, these sets are approximately equal in size and MATH. |
math/0106100 | CASE: Immediate from REFEFf. CASE: Immediate from REFEFb. |
math/0106100 | (See below.) |
math/0106100 | CASE: Let a run of MATH be a maximal consecutive subset of MATH. (So MATH has only REF-membered runs, while MATH has only REF-membered runs.) So, MATH is the first element in the MATH-th run of MATH and MATH is the first element after the MATH-th run of MATH. CASE: We know from REF that MATH or MATH. But the disjoint union of MATH with MATH or MATH, respectively, yields MATH or MATH. So REF follows from MATH. CASE: We need only prove REF : MATH . Informally: MATH first becomes greater than MATH for MATH, since MATH because MATH. Throughout the first run of MATH, MATH maintains its lead, losing it only at the least MATH for which MATH (since MATH, so MATH). This pattern repeats during successive runs of MATH. |
math/0106100 | Without loss of generality, we can suppose there is a set, MATH, such that MATH and MATH. By REF, MATH and MATH . |
math/0106100 | CASE: Let MATH. Since neither MATH nor MATH outpaces the other, MATH is infinite. By REF, let MATH be a non-principal ultrafilter which contains MATH. Then MATH. CASE: If MATH is an alternating pair, so is MATH. So, by REF there is a MATH such that MATH. But then MATH. |
math/0106100 | Recall that every infinite completion of MATH is equivalent to MATH for some total and congruous remainder function, MATH. (See REFEF.) Given such a MATH, let MATH for each MATH. Then REF holds for each MATH: MATH . Let MATH. The intersection of any finite subset, MATH, of MATH is infinite. If MATH is a finite subset of MATH, then MATH, where MATH is some finite subset of MATH. So MATH. But MATH is congruous, so the restriction of MATH to the finite domain MATH has infinitely many solutions (see REFEFa). Since the intersection of any finite subset of MATH is infinite, there is a non-principal ultrafilter MATH such that MATH. By REF , MATH, so MATH. |
math/0106100 | Suppose that MATH is a linear ordering under which MATH forms a MATH-sequence. Then we could define MATH-outpaces MATH as follows: MATH and define the principle: MATH. If MATH, then MATH. Modifying REFEF, we could produce standard models of MATH over MATH which satisfy MATH and these will not, in general, satisfy MATH. Suppose, for example, that MATH is the ordering: MATH where MATH is the set of prime numbers and MATH is its complement. In any model of MATH, MATH and MATH will be nearly the same size, as the evens and the odds are in normal outpacing models. But the evens are much smaller than MATH, so MATH and MATH is false in MATH models. |
math/0106100 | CASE: Let MATH. So MATH contains all numbers between REF, between REF, between REF, and so forth. If MATH, then MATH and MATH. So MATH cannot have a limit. CASE: Suppose MATH is given. Construct the set MATH as follows: MATH CASE: Modify the construction for REF in the obvious ways. |
math/0106100 | MATH . |
math/0106100 | Since MATH, MATH outpaces MATH. But if MATH, then MATH outpaces MATH. So, MATH. |
math/0106100 | CASE: Immediate from REF . CASE: Let MATH be an infinite set with density MATH, let MATH be an infinite subset of MATH, where MATH, and let MATH. There are uncountably many subsets of MATH with distinct densities in MATH, though MATH, for any MATH. Suppose now that MATH, MATH, and MATH. Then MATH, by REF so MATH, by MATH. But MATH, since MATH was obtained by removing from MATH a set with density REF relative to MATH. |
math/0106100 | CASE: Suppose that MATH and MATH. By MATH, there are two sets, MATH and MATH, for which MATH . MATH CASE: Suppose that MATH and MATH. If MATH or MATH, then MATH, so MATH . To apply REF , we need to find two sets, MATH and MATH such that MATH and MATH . Assuming that MATH and MATH, let MATH and MATH . Then MATH, by MATH, and MATH, by MATH. Since adding or removing a single element has no effect on the density of a set, MATH . So, by REF , MATH. |
math/0106100 | Let MATH be a set with density MATH and let MATH be a divergent set which neither outpaces nor is outpaced by MATH. Let MATH be MATH is infinite, so MATH is a member of some non-principal ultrafilter, MATH. MATH, so REF is false in MATH. |
math/0106101 | If MATH is a conformal foliation then it is easy to see that, locally, we have MATH for any local dilation MATH of MATH (see, for example, CITE). From REF and CITE , it follows that a conformal foliation MATH produces harmonic morphisms if and only if, in the neighbourhood of each point, there exists a local dilation MATH such that MATH . Therefore, for any such local dilation MATH , we have MATH and the proof follows. |
math/0106101 | Suppose that MATH , and let MATH be a submersive harmonic morphism whose fibres are open subsets of leaves of MATH . Let MATH be local coordinates on MATH given by harmonic functions (see CITE or CITE for a proof that such local coordinates exist). Then, because MATH is a harmonic morphism, MATH is a submersive harmonic map. From the fact that MATH is real-analytic it follows that the components of MATH are real-analytic functions on MATH , and hence MATH is locally defined by real-analytic submersions. Thus MATH is a real-analytic foliation. If MATH , the proof is ended. Suppose that MATH . From the fact that both MATH and MATH are analytic it follows that the one-form MATH is analytic and hence the dilation of any harmonic morphism produced by MATH is an analytic function. Let MATH be a submersive harmonic morphism produced by MATH . We shall show that MATH is real-analytic in harmonic coordinates. Indeed, since the dilation MATH of MATH is analytic and MATH we have that, in harmonic coordinates, MATH has analytic components. Then, the analyticity of the atlas MATH on MATH given by the harmonic coordinates follows from the analyticity of harmonic functions on analytic Riemannian manifolds (applied to the charts of MATH). |
math/0106101 | Let MATH be defined by MATH . Obviously MATH where MATH ; suppose that MATH is unbounded. For each MATH , let MATH be such that MATH for every MATH. Because MATH is unbounded we must have that MATH is unbounded. By passing to a subsequence, if necessary, we can suppose that MATH . Because MATH we have MATH. But MATH where MATH , a contradiction. Hence MATH must be bounded. |
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