paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0009236 | To check that the first pairing is well defined, let MATH be a coboundary in MATH. Then MATH is also a coboundary and we see that MATH since MATH is MATH-invariant and therefore MATH. Let MATH in MATH, where MATH and MATH. Then, by REF , MATH, so there exists a MATH-invariant invertible element MATH such that MATH. The... |
math/0009236 | We prove that the following maps define MATH-isomorphisms between MATH and MATH inverse to each other: MATH where MATH is the MATH-matrix of MATH. Since MATH and MATH and MATH therefore MATH, where MATH. Since MATH where MATH, we see that MATH is an algebra map. Also since MATH and MATH, we conclude that MATH which sho... |
math/0009236 | We prove that the following maps define a MATH-isomorphism between MATH and MATH inverse to each other: MATH . Since MATH and MATH we obtain MATH. Now since MATH is an algebra map. To check MATH preserves the MATH-actions, we see that MATH . |
math/0009239 | The sufficiency of the condition follows from the fact that MATH is a NAME subalgebra when MATH is graded. In order to check the necessity of the condition, notice that MATH gives a NAME system allowing to compute the homogeneous components of a vector field MATH. |
math/0009239 | Using NAME identity, we check that MATH and consequently MATH by induction on MATH. By definition, MATH for all MATH. Then, we check, by induction on MATH, that MATH. Therefore, MATH is a NAME subalgebra. It is trivially the smallest one to contain the subspaces MATH, MATH and MATH. |
math/0009239 | It is obvious that MATH. Furthermore, if MATH or MATH, MATH . Now, if MATH, then, for every polynomial function MATH, the field MATH belongs to MATH. |
math/0009239 | Notice that MATH cannot be made only of constant and linear vector fields. Indeed, it would then be included in the maximal subalgebra REF presented in the introduction, for instance. Therefore, MATH for some MATH. The conclusion follows from REF . |
math/0009239 | Let MATH be a stable subspace of MATH under the action of MATH. The space MATH satisfies the hypotheses of REF . Its algebraic closure is an infinite dimensional proper subalgebra containg MATH properly, hence a contradiction. Let now MATH be a non trivial ideal of MATH. It contains at least one constant vector field s... |
math/0009239 | Indeed, MATH. The requirement for MATH to intertwine the action of MATH on MATH precisely means that MATH. If MATH and MATH, the equalities MATH show that MATH. The inclusions are strict because the dimension of MATH is infinite and because the dimension of MATH, for all MATH, is strictly less than that of MATH. |
math/0009239 | The sufficiency of the condition is obvious. Notice that a complex structure on MATH stabilizes the eigenspaces of MATH. Let MATH be a complex structure on MATH. Let MATH be the adjoint of MATH with respect to the MATH form MATH of MATH, that is, MATH . The so defined MATH intertwines the action of MATH on MATH. Moreov... |
math/0009240 | : This is an elementary exercise in modal reasoning. Scheme REF is actually a special case of scheme REF, obtained by replacing MATH with MATH in REF to obtain MATH. This is just MATH because MATH is equivalent to MATH. Conversely, scheme REF implies scheme REF because MATH implies MATH. Scheme REF is the contrapositiv... |
math/0009240 | : Let MATH simply be the statement, ``MATH is hereditarily countable." Of course, this is forceably necessary, since any set can be made countable by forcing, and once countable, it can never be made uncountable again by further forcing. But by the hypothesis on MATH, the statement MATH is not true. |
math/0009240 | : If a sentence MATH is forceably necessary over a forcing extension, then it was already forceably necessary over the ground model, and if necessary in the ground model, it remains necessary in any forcing extension. That is, MATH is downward absolute from a forcing extension, and MATH is upward absolute to any forcin... |
math/0009240 | : In order to highlight various aspects of the theory, I will actually give two proofs of this theorem. The first proof is elementary, but the second proof will generalize to MATH. We begin with a simple observation. For no sentence MATH are both MATH and MATH forceably necessary. : Suppose that MATH is forced to be ne... |
math/0009240 | A Second NAME Second Proof: I would like now to give an alternative proof of REF , relying on a more traditional iterated forcing construction. It is this iterated forcing argument that will generalize to the case of MATH. Let me motivate the first lemma by mentioning that if there is an inaccessible cardinal MATH, the... |
math/0009240 | : I claim, first, that if there is a model of zfc, then there is one in which the definable ordinals - and I mean ordinals that are definable without parameters - are unbounded. To see this, suppose that MATH is a model of MATH, not necessarily transitive, and let MATH be the cut determined by the definable elements of... |
math/0009240 | : REF shows that every model of MATH has a forcing extension that is a model of MATH. Conversely, suppose that MATH is a transitive model of MATH. Consider the MATH of MATH, which must have the form MATH, for MATH. The arguments of the previous theorem establish that every ordinal that is definable in MATH is countable... |
math/0009240 | : First, one shows that if there is a model of zfc, then there is a model of MATH is uncountable. This can be proved in the same way as REF , by simply adding the assertion MATH to the theory. The NAME Reflection Theorem shows that any finite collection of formulas is absolute from MATH to MATH for a closed unbounded c... |
math/0009240 | : I will prove the theorem with a sequence of lemmas. Every model of theory REF has an inner model of theory REF. Specifically, if MATH holds, then MATH is inaccessible in MATH, and MATH. : Assume MATH and let MATH. I claim, first, that MATH is inaccessible to reals. To see this, suppose MATH is a real, and let MATH be... |
math/0009240 | : Over any model one can force ch without adding reals, and this will preserve MATH. Alternatively, the model of REF satisfies MATH, since it is obtained via the NAME collapse of an inaccessible cardinal. To obtain a model of MATH, we will simply add NAME reals over this model. Specifically, I claim that the model MATH... |
math/0009240 | : Suppose that MATH is consistent with the existence of a proper class of, for example, measurable cardinals. The argument of REF produces a model of MATH with a proper class of measurable cardinals. Since the forcing of REF has size at most MATH, the measurable cardinals above MATH survive to the forcing extension in ... |
math/0009240 | : Suppose that the Maximality Principle holds, but that the inaccessible cardinals are bounded. Then the assertion ``there are no inaccessible cardinals" is forceably necessary, because one could force, in a permanent way, to make all the inaccessible cardinals countable. Thus, under mp, there must have been no inacces... |
math/0009240 | : It suffices, by REF, to find a violation of covering. For this, it suffices to show that MATH fails to compute the successors of singular cardinals correctly. In fact, I will show that MATH fails to compute the successor of any cardinal correctly. Consider any cardinal MATH and its successors MATH and MATH in the two... |
math/0009240 | : Clearly, REF is implied by REF, which is in turn implied by REF. Let me argue that REF is implied by REF. If MATH, then clearly MATH is a strong limit in MATH, and hence, if regular, MATH is inaccessible in MATH. Thus, the inaccessible cardinals are unbounded in MATH, and hence also in MATH, so statement MATH holds. ... |
math/0009240 | : By this, I mean that such a class MATH will not compute successor cardinals above MATH correctly, that every cardinal above MATH will be a limit of inaccessible cardinals in MATH, and so on, and that every set MATH definable in MATH from an object in MATH for some MATH above MATH will be in MATH. Suppose, therefore, ... |
math/0009240 | : Fix any set MATH, and let MATH be a real collapsing cardinals so that MATH can be coded by a real MATH in MATH. Let MATH. This is definable from MATH, and is invariant in any further extension of MATH. Thus, by the previous corollary, if MATH holds, then MATH does not compute the successors of singular cardinals abov... |
math/0009240 | : Assume MATH. I will show, by induction on MATH, that every MATH formula MATH is absolute to any set forcing extension. This is trivial for MATH (and indeed, it holds automatically up to MATH by the NAME Absoluteness Theorem). Assume by induction that it is true for all MATH formulas. Further, since MATH holds in all ... |
math/0009240 | : CITE has shown that projective absoluteness is equiconsistent with the existence of infinitely many strong cardinals. |
math/0009240 | : CITE has proved that if there is no inner model of a NAME cardinal and if the universe is closed under sharps and MATH is ineffable, then there is a definable class MATH, now widely known as the NAME core model, which is invariant under set forcing and which computes the successors of singular cardinals correctly. Si... |
math/0009240 | : I will first prove that if there are a proper class of NAME cardinals, then the Maximality Principle MATH holds for local assertions using real parameters in the ground model. From this, since the hypothesis of a proper class of NAME cardinals is preserved by set forcing, it follows that the same conclusion holds in ... |
math/0009240 | : In the proof of REF , let us simply ensure that the forcing MATH at stage MATH comes from MATH. Then, define MATH to be the finite support or countable support iteration of the MATH, depending on the closure of MATH. Having done so, the iteration MATH, as well as the tail forcing MATH, will be in MATH, and so the arg... |
math/0009240 | : This argument follows the first proof of REF . Suppose that MATH is any model of zfc, and let MATH be the collection of sentences that are forceably necessary over MATH, referring throughout this argument only to forcing notions in MATH. I claim first that the theory MATH is consistent. Given any finite subcollection... |
math/0009240 | : The corresponding classes of posets have the closure properties of the theorem. |
math/0009240 | : Clearly, the first theory directly implies the second theory. Furthermore, I have already shown that the third and fourth theories are equiconsistent. Every model of theory REF contains an inner model of theory REF. This Lemma is a consequence of the following Lemma: If MATH holds (in the language with ordinal parame... |
math/0009240 | : Using the techniques of REF, one can show that if the first theory holds, then MATH is inaccessible in MATH and MATH. And this has already been proved equiconsistent with the second and third theories. Conversely, the second and third theories are equiconsistent with MATH for an inaccessible cardinal MATH. Given this... |
math/0009242 | We will show that for MATH this small, on average MATH increases at each step. If MATH, then the size of MATH goes up by MATH, but if MATH, then the size of MATH may decrease by at most MATH [removing MATH (not previously included), MATH, and some neighbors]. Hence MATH which is positive precisely when MATH. Given an i... |
math/0009242 | Let MATH, and suppose that MATH. Let MATH be the event that MATH and MATH. Then MATH where the last step is exactly our assumption. Note that neither MATH nor MATH depends on MATH. Hence, summing over MATH, MATH . This completes the proof. MATH . |
math/0009242 | The acceptance probabilities were chosen precisely to match the requirements of REF . For instance, with heat bath RR, the left side of the equation in REF equals MATH which reduces to the right side of the equation with MATH. The calculation for arbitrary RR is entirely similar. MATH . |
math/0009242 | We use a potential function that rewards us for adding edges and penalizes us for connecting components. Let MATH where MATH will be determined later. When the edge MATH we attempt to add to MATH is between two vertices already connected in MATH, then MATH always goes up by REF, making this case uninteresting. It is wh... |
math/0009244 | From REF and MATH we see that MATH are the eigenfunctions of MATH. Then the theorem is obtained from REF since it implies that the range of MATH is dense. |
math/0009244 | The symmetry of the operator MATH is trivial. Then we deduce that MATH is essentially self-adjoint on MATH. |
math/0009244 | For MATH, we have MATH due to MATH. Then MATH which implies that MATH is one-to-one and MATH. Similarly we have MATH and hence MATH. |
math/0009244 | We define the functions MATH and MATH by MATH . Then we see that MATH and MATH are real-holomorphic. Equivalently, MATH and MATH are real-holomorphic. By REF , MATH or the matrix MATH is invertible, which implies MATH is real-holomorphic. |
math/0009244 | Since the operator MATH is expressed in terms of MATH and MATH, it is enough to show the MATH cases. The NAME group of type MATH acts on the space of function on MATH by the permutation of the variable. We denote the action of MATH on MATH by MATH. Let us recall the definition of MATH, that is, MATH . Let MATH be a fun... |
math/0009244 | We prove MATH. For the other cases, the proofs are similar. From the definition, the function MATH is symmetric and rational with respect to the variables MATH. The possible poles of the rational function MATH are MATH and the degree of each pole is one, but it contradicts to the NAME group invariance of the function M... |
math/0009244 | Since MATH and MATH, it is enough to show MATH. We have the equality MATH for some non-zero constant MATH, which follows from the correspondence between the integration of the MATH-invariant function and the one of the MATH. From this equality, REF , and the commutativity MATH, if we show MATH where MATH and MATH, then... |
math/0009244 | It is trivial for the MATH case. We assume MATH. Let MATH. Then MATH is a polynomial in the parameter MATH of degree at most MATH and MATH. We set MATH. Then MATH, because MATH for all MATH. For MATH and MATH, we set MATH and MATH. We fix the functions MATH. It is enough to show that the equations MATH hold for MATH an... |
math/0009244 | We introduce variables MATH and set MATH . The numbers MATH and MATH are given in advance. We will investigate the conditions for the coefficients of the formal power series MATH and MATH satisfying the following relations MATH . We set MATH . By comparing the coefficients of MATH, we obtain that REF are equivalent to ... |
math/0009244 | The finiteness of the summation MATH follows from REF and the fact that the NAME polynomial forms a basis of MATH. The non-degeneracy of the joint eigenvalues MATH follows from the non-degeneracy of the joint eigenvalue of the NAME polynomial. |
math/0009244 | Let MATH be the sum of all divisors of MATH. By REF , we have MATH. Since the convergence radius of the series MATH is equal to MATH, the convergence radius of the series MATH is equal to or less than MATH. Therefore we have the lemma. |
math/0009244 | It follows from REF and the inequality MATH . |
math/0009244 | From REF , we have MATH where MATH is the highest root of the root system MATH. Since MATH and MATH, we have the lemma. |
math/0009244 | This follows from the NAME formula (CITE, p. REF.). |
math/0009244 | It is sufficient to show MATH for MATH and MATH. This inequality is equivalent to MATH. From the property MATH, we have MATH. |
math/0009244 | Immediate from the equality MATH. |
math/0009244 | First, we expand MATH by using NAME REF. Then MATH is expressed as the linear combination of MATH, where MATH and MATH. We set MATH . We repeatedly apply NAME REF for MATH. Then MATH is expressed as the linear combination of MATH, where MATH for some MATH. From NAME REF, we have MATH. Applying NAME REF for MATH and MAT... |
math/0009244 | Since the normalized NAME polynomials form the complete orthonormal system with respect to the inner product MATH, we have MATH. We fix MATH. Let MATH be the smallest integer which is greater or equal to MATH. If MATH, then we have MATH by REF and the orthogonality. Therefore we have MATH . Similarly, we have MATH. Sin... |
math/0009244 | Let us recall that the operator MATH is defined by the NAME series REF . We fix the number MATH and set MATH. From the expansion REF , there exists a number MATH such that the inequality MATH holds for MATH REF and MATH. In this case, we have MATH. We write MATH. For the series MATH, write MATH. We have MATH for each M... |
math/0009244 | Since the spectrum MATH is discrete, there exists a positive number MATH such that MATH. We write MATH. From REF , we obtain that for each MATH, there exists MATH and MATH which does not depend on MATH such that MATH for all MATH such that MATH and MATH. Let MATH be the length of the circle MATH and write MATH. By inte... |
math/0009244 | Since MATH, it is enough to show that the function MATH is holomorphic when MATH is sufficiently small and MATH. We count roughly the number of the elements of MATH of a given length. The rough estimate is given by MATH. We will use this in REF . In the proof, we will use the notations and the results written in REF. I... |
math/0009245 | Let MATH be a small perturbation of a critical point MATH, such that MATH and MATH. The derivative of the function MATH, at MATH, is given by MATH where MATH. If for all MATH, we have that MATH then MATH . The second equation is obtained by considering a smooth curve MATH given by MATH, where MATH. It follows from the ... |
math/0009245 | Since MATH we can perform the computation of MATH by fixing a class of MATH. For a class MATH, we fix a map MATH representing MATH and MATH. Thus, MATH . However, MATH . Let MATH. By NAME 's formula, MATH and MATH. Consequently, MATH . The group MATH is described below in REF ; (the symbol MATH stands whether the group... |
math/0009245 | The Minimax Principle implies that the minimum value is attained in all connected component of MATH. REF shows that it is possible to construct non-contractible families of elements in MATH, consequently, by applying the Minimax Principle, it follows that there exist stable and non-stable critical points. In other word... |
math/0009245 | The proof is splited in two easy claims; CASE: If MATH, then MATH. Let MATH; MATH CASE: If MATH, then there exits MATH and MATH such that MATH . Let MATH and MATH; MATH . |
math/0009245 | Let MATH; so MATH. NAME, MATH this implies that MATH and MATH . |
math/0009245 | Its follows from the discussion above, since it was concluded that MATH . |
math/0009246 | We prove this proposition by using the weak compactness REF . If the compactness fails, then there exists a finite number of bubble points MATH such that MATH has a non-trivial area concentration in each bubble point (we follow notations in REF above). Consider two cases: MATH or MATH is a torus or higher genus. In the... |
math/0009246 | Write MATH . Note that the NAME energy, NAME energy and area are uniformly bounded along this sequence of metrics. Then we have MATH and MATH . Here MATH is some uniform constant in this proof, and its value may change from line to line. Since area is fixed along the flow: MATH . It follows that MATH . Since the underl... |
math/0009246 | Suppose MATH . Then MATH in MATH . Thus MATH in MATH for any MATH . In other words, MATH uniformly. The later in turn implies that first eigenvalues of MATH must converge to that of MATH . The last statement of the proposition just follows from the NAME inequality. |
math/0009246 | Since the NAME flow exists for long time, REF holds for time MATH for some constant MATH . Therefore there exists at least a sequence of numbers MATH such that MATH . REF implies that there exists a subsequence (denoted again by MATH), a subsequence of conformal transformations MATH and a constant scalar curvature metr... |
math/0009246 | Let MATH and let MATH denote any generic constant. Then MATH as MATH . We want to show that MATH decays exponentially fast (note that MATH in the following calculation): MATH . Now we need to consider two cases: MATH or MATH . The first case is easier, while the second case is more delicate. Consider the first case. If... |
math/0009246 | We already know that there exists a family of conformal transformations MATH such that MATH converges to a constant scalar curvature metric MATH . Then the set MATH must be compact . Otherwise, suppose that MATH is a non-compact family of conformal transformations. By a direct calculation, one should yield: MATH . On t... |
math/0009249 | We shall write MATH and MATH; we are going to view MATH as the pre-Hilbert MATH-module described above, and also as a dense subspace of the NAME MATH-module MATH. For MATH, we define MATH by MATH . We shall prove that MATH extends to a unitary operator of MATH onto MATH which intertwines the given representations. We f... |
math/0009249 | Every regular representation MATH induced from a faithful representation MATH of MATH has the same kernel MATH, and MATH has kernel MATH. We choose MATH to be the restriction MATH of a faithful nondegenerate representation of MATH. Then MATH is the restriction of the regular representation MATH of MATH, and we can appl... |
math/0009249 | This is the special case of REF in which MATH; the dynamical system MATH is then trivially amenable. |
math/0009249 | REF immediately gives the ``if" direction. So suppose MATH is equivalent to the regular representation MATH for some representation MATH of MATH. Let MATH, and note that MATH is equivalent to MATH. REF implies that MATH is equivalent to MATH, and applying MATH shows that MATH is equivalent to MATH. |
math/0009249 | This follows from REF by taking advantage of the symmetry of the situation. Suppose that MATH is a representation of MATH instead. Let MATH and note that MATH is regular because it is equivalent to MATH. By REF MATH is the restriction of a covariant representation MATH of MATH. |
math/0009249 | We can use a partition of unity on MATH to write every function in the dense subalgebra MATH as a sum of functions supported on MATH-saturated open subsets of MATH which are trivial as MATH-bundles. Since MATH is nondegenerate and in particular nonzero, MATH must be nonzero on one of these sets. More formally, there is... |
math/0009249 | From REF and a NAME 's Lemma argument, we obtain a decomposition MATH into MATH-invariant subspaces, each of which admits a suitable nondegenerate representation MATH. Now we just take MATH, and apply NAME 's imprimitivity theorem as described above. |
math/0009249 | To avoid having to write out the opposite version of REF , we instead prove the equivalent assertion that the regular representation MATH of the system MATH satisfies MATH. To do this, we apply REF with MATH absent. Then MATH is the bimodule MATH of CITE, MATH is the trivial bimodule MATH, and REF says that MATH . Sinc... |
math/0009249 | Suppose that MATH is amenable. Applying REF to MATH shows that MATH . Since MATH, MATH factors through a faithful representation MATH of MATH; the amenability of MATH implies that MATH is faithful, or, equivalently, that MATH. REF says that the NAME correspondence MATH carries MATH to MATH. Thus MATH . On the other han... |
math/0009250 | We use induction on MATH; the case MATH follows directly from hypothesis PL REF . Suppose the result is true for MATH, and fix MATH, MATH with property MATH and let MATH be a tree on MATH isomorphic to MATH such that every MATH-subsequence of MATH satisfies MATH. Let MATH be the sequence of initial nodes of MATH with M... |
math/0009250 | Let MATH, and consider the following two possibilities: CASE: MATH; CASE: MATH. In REF so that MATH . For REF MATH and REF now follows from REF . |
math/0009250 | We may assume that MATH for some sequence MATH. We may also assume that MATH with MATH isomorphic to the replacement tree MATH. It is sufficient to find a sequence MATH and trees MATH of order MATH such that MATH satisfies the conditions of the lemma for each MATH. Let MATH and let MATH and sets MATH satisfy for MATH: ... |
math/0009250 | Let MATH be the given MATH tree on MATH of order MATH and let MATH be a MATH tree on MATH of order MATH. Let MATH be the sequence of terminal nodes of MATH, so that MATH. Choose MATH, MATH and for each MATH find MATH such that MATH whenever MATH. Apply the previous lemma to MATH for MATH and MATH with MATH to obtain a ... |
math/0009250 | If MATH, then, since the closure of a MATH tree is a MATH tree for some MATH by REF , it follows from REF that there exists an infinite sequence as in the statement of the theorem. Otherwise we assume the index is countable and let T be a MATH-tree on MATH of order MATH. By the previous lemma there exist numbers MATH a... |
math/0009250 | We only need make a couple of modifications to the proof of REF . Instead of embedding MATH into MATH and using a basis there we use the basis MATH of MATH. Then in the proof of REF we must ensure that we construct a block basis tree. But this is easy. In REF , for each terminal node MATH we choose MATH so that MATH, a... |
math/0009250 | REF are results of CITE and CITE stated in terms of the MATH-index. In particular REF follows from REF or CITE Corollary of REF , while REF follows from CITE and CITE. Before we can give the proof of REF we need some results on the relationship between MATH and MATH sequences, so we shall postpone the proof until we ha... |
math/0009250 | We prove this by induction on MATH. There is nothing to prove for MATH, and if the result has been proven for every MATH, then it is also clear when MATH is a limit ordinal. Thus suppose the result has been proven for MATH, let MATH, and let MATH be a tree of order MATH. By taking a subtree we may assume that MATH has ... |
math/0009250 | We shall show that for each MATH there exists a MATH-tree on MATH of order MATH if and only if there exists a MATH-tree on MATH of order MATH. Let MATH and let MATH be a MATH-tree on MATH of order MATH for some MATH. Choose MATH and a sequence MATH decreasing rapidly to zero. Let MATH be the tree obtained when REF is a... |
math/0009250 | From REF we know that MATH, and so it suffices to show (see for example, Monk CITE) that if MATH, then MATH. We may regard MATH as a subspace of MATH, and let MATH be a monotone basis for MATH. Let MATH be a MATH-weakly null tree on MATH of order MATH. To make a tree of order MATH we want to add a tree of order MATH af... |
math/0009250 | We first show by induction on MATH that if MATH is a MATH-block basis tree on MATH of order MATH, then we can extract a certain subtree MATH isomorphic to MATH. The tree MATH will have the property that each MATH-subsequence MATH is a normalized block basis of the shrinking basis MATH and hence is weakly null. Thus MAT... |
math/0009250 | As usual we use induction on MATH. For the initial case MATH, let MATH and let MATH be a weak-MATH relatively open neighborhood of MATH. We may find MATH such that MATH and MATH. Set MATH, then MATH is the required tree. We next suppose the result has been proven for MATH; let MATH, and let MATH be a weak-MATH relative... |
math/0009250 | Let MATH and let MATH be the tree for MATH from the previous lemma for MATH. Replace each node MATH with the node MATH to obtain the tree MATH which is still isomorphic to MATH. Clearly, if MATH is any MATH-subsequence, then MATH and MATH. Let MATH have property MATH if it is weak-MATH null with MATH, and property MATH... |
math/0009250 | Let MATH, to be chosen later. We first choose a sequence MATH with MATH for each MATH. Since MATH is separable we may assume MATH is weakly NAME (by taking a subsequence of MATH and then the same subsequence of MATH). Again, by taking subsequences and using that MATH, we may assume that MATH if MATH (where MATH is smal... |
math/0009250 | We have that MATH and we want to construct a MATH-weakly null tree on MATH of order MATH and constant MATH. Let MATH be the tree on MATH for some MATH from REF . We want to construct a tree MATH in MATH, isomorphic to MATH, so that if MATH has immediate successors MATH etc., and MATH are the corresponding nodes of MATH... |
math/0009250 | As usual we proceed by induction on MATH. If MATH, then MATH where MATH and MATH is a normalized weakly null sequence. For each MATH pick MATH with MATH, then choose a subsequence MATH which converges weak-MATH to some MATH. Choose a sequence MATH; since MATH is weakly null, it follows that for each MATH there exists M... |
math/0009250 | If MATH is a MATH-weakly null tree on MATH of order MATH, then there exists a branch functional for each branch of MATH. Thus we may take MATH in the above lemma, to obtain MATH. |
math/0009250 | We first show that MATH is a monotone basic sequence, and then apply the NAME theorem to obtain that its span is all of MATH. Let MATH, then MATH . Since MATH is hereditary and the enumeration of MATH is admissible, it follows that there is an index MATH and MATH such that MATH. Next, observe that for all MATH with MAT... |
math/0009250 | The heart of the proof lies in choosing an appropriate block basis of MATH. To do this we shall construct a tree isomorphism MATH from MATH into MATH by induction, and then the map from the basis MATH of MATH to a block basis of MATH will be given by MATH . This immediately gives the ordering requirement on MATH that i... |
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