What is Zariski Topology?
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The Zariski topology is a topology on R^n (more generally on F^n for some field). The closed sets are defined to be sets of the form {x in R^n: p(x)=0} where p is a polynomial. That this does define a topology follows from properties of polynomials. On R, this is exactly the co-finite topology, but on R^n with n>=2, this is not the case. The main difficulty with this topology is that it is never Hausdorff, so limits need not be unique. It does reflect a lot of the geometry of algebraic varieties, though.
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Suppose a continuous open map f: R -> R is not monotonic.  Then we may assume, without loss of generality, that there are points x < y < z in R such that f(y) is the maximum (or minimum) element of f((x,z)).  This implies that f((x,z)) is not open, contradicting the assumption that f is open.
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Domain:  all real numbers\nRange:  [2, positive infinity]   (all real numbers from 2 to positive infinity)
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It has dimension 2.\nI will not bother to prove that this subset is indeed a subspace except to note that (0,0,0) is an element.\nThis is most readily seen by constructing a basis.\nI propose that {(1,1,0),(0,1,1)} is a basis for this space.\nLet (a,b,c) be an element of this space. Then (a,b,c)=(a,a+c,c) =  a*(1,1,0) + c*(0,1,1)\nThis our proposed basis Spans this subspace.\nNow I think we can clearly see that our proposed basis is linearly independant since:\nIf:   a*(1,1,0) + c*(0,1,1) = (0,0,0)\nthen: 0 = a, 0 = a+c and 0 = c by equating the components.\nThus we have a linearly independant spanning set.\nThus we have a basis!\nSince our basis is of magnitude 2, our space must be of finite dimension 2.
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What's so special about the Mandelbrot set? Its boundary exhibits complicated structure at all length scales from about 1 down as far as you wish (or as you computer will take you) to zero. In other words, it is a fractal . You can experiment with this set yourself using the applet (the implementation of a Java code on a web page) on this page. To do so you need to have a Java compatible browser such as a recent version of Netscape, \n\nWhat is the Mandelbrot set? It's the the set of all complex numbers z for which sequence defined by the iteration \n\nz(0) = z,    z(n+1) = z(n)*z(n) + z,    n=0,1,2, ...    (1) \nremains bounded. This means that there is a number B such that the absolute value of all iterates z(n) never gets larger than B. A bounded sequence may or not have a limit. For example, if z=0 then z(n) = 0 for all n, so that the limit of the (1) is zero. On the other hand, if z=i ( i being the imaginary unit), then the sequence oscillates between i and i-1, so remains bounded but it does not converge to a limit. \n\nYou may ask, what's so special about the particular iteration (1), and why do we use complex numbers instead of real ones. In a sense, the formula (1) is the simplest other than a linear formula which would give rise to a much simpler and quite uninteresting picture. (The analog of the Mandelbrot set would be empty or the entire plane.) If we restricted the iteration (1) to the real instead of complex numbers then again we would get an uninteresting picture: the interval from -2 to 0. \n\nMuch of the fascination of the Mandelbrot set stems from the fact that an extremely simple formula like (1) gives rise to an object of such great complexity.
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There are two interpretations I can see here. The first is that the homology groups of the 0-dimensional sphere are needed. For this, the answer is trivial: you just get Hn(pt)+Hn(pt) (direct sum). Over a ring R, Hn(pt)=R if n=0 and is 0 if n>0. So the answer is R+R for n=0 and 0 otherwise.\n\nIf the asker actually means SO(k), the special orthogonal group, this is much trickier. It is *not true* that SO(k) is homeomorphic to S^(k-1) (this only happens if k=2; for k=3 you get a projective space). In fact, the dimension of the manifold SO(k) is k(k-1)/2. In this case, I don't know the answer off the top of my head.
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The term "isolated singularity" just refers to a singularity, z_0, in which you can create a neighborhood around it where f is analytic throughout that neighborhood with the exception of the point z_0.  Visualize this as just a single point in the plane.  If this point was a limit point of some series of singularities then it would be very hard to work with, indeed - that's why we want all our singularities to be isolated singularities!\n\nWell, to begin to make sense of your question we first need to know what a Laurent expansion is:  Well, you should be very careful about putting this together, but sloppily said this is an expansion of the form: \nf(z) = Sum(nfrom -infinity to +infinity)a_n(z-z_0)^n\n\nBut what in the world does this have to do with singularities?  Well, if a_n is 0 for all negative n, then we have found a power series expansion for the function - this it is (almost) analytic.  ie, it can be made analytic by "properly" defining f(z_0).  This is the exact notion behind removable singularities - they can be removed!\n\nHow about if all but a finite number of a_n's are zero when n is negative?  Then we could multiply by a large enough value of (z-z_0)^n to make a new function which has a removable singularity.  This is what we call a pole.  It is not removeable, but it can still be worked with, in a way...\n\nFinally if infinitely many a_n with n negative are nonzero then there is no way to make this thing behave like an analytic function near the point z_0.  It is an essential singularity.  A cool result from Picard's theorem says (roughly)that any neighborhood of an essential singularity is mapped to the entire complex plane (except possibly one point).
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Karnaugh map:\n\nCheck out the following link which will help.
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Two and a half centuries later, Georg Cantor made Galileo’s paradox a basis for a mathematical theory of infinity. Cantor, who lived from 1845 to 1918, was a Russian-born German mathematician with an artistic streak and a keen interest in theology. He saw that the collapse of the familiar logic of part and whole yielded a new definition of infinity, one that did not rely on the vague notion of endlessness. An infinite set, as Cantor characterized it, is one that is the same size as some of its parts—a set, then, that can lose some of its members without being diminished. Now Cantor was in a position to ask a novel question: Are all infinities the same, or are some more infinite than others?\n\nLooking for an infinity bigger than that of the whole numbers, he started by considering the set of fractions. It seemed a good bet, since fractions are densely ordered on the number line: between every two whole numbers there are infinitely many fractions. (Between 0 and 1, for example, lie 1/2, 1/3, 1/4, 1/5, and so on.) Yet Cantor, to his surprise, was able to find an easy way of matching up the whole numbers and the fractions one to one. Despite initial appearances, these two infinities turned out to be the same. Perhaps, he thought, all infinite sets, by dint of being inexhaustible, were of the same magnitude. But then he looked at the “real” numbers—those marking off the points on a continuous line. Could they, too, be paired up, one to one, with the whole numbers? By a surpassingly clever bit of reasoning, called the “diagonal proof,” Cantor showed that the answer was no. In other words, there were at least two distinct infinities, that of the whole numbers and that of the continuum, and the second was greater than the first.\n\nBut was that the end of it? In searching for a still larger species of infinity, Cantor looked to higher dimensions. Surely, he thought, there must be more points in a two-dimensional plane than on a one-dimensional line. For a couple of years, he struggled to prove that the points in a plane could not be paired off one to one with the points in a line, only to find, in 1878, that such a correspondence was indeed possible. A simple trick showed that there were exactly as many points on a one-inch line as there were in all of space. “I see it, but I do not believe it!” Cantor wrote to a colleague.\n\nWith the discovery that neither size nor dimension was the way to higher infinities, the quest seemed to stall. But after more than a decade of intense work (interrupted by a stay in a sanatorium after a nervous breakdown), Cantor arrived at a powerful new principle that enabled him to resume the ascent: there are always more sets of things than things. This is obvious enough in a finite world. If you have, say, three objects, you can form eight different sets out of them (including, of course, the empty set). Cantor’s genius lay in extending the same principle into the realm of the infinite.\n\nTo make matters a little less abstract, you might pretend that we live in a world with an infinite number of people. Now consider all the possible clubs (sets of people) that might exist in this world. The least exclusive of these clubs—the universal club—will be the one of which absolutely everyone is a member. The most exclusive—the null club—will be the one that has no members at all. In between these two extremes will lie an infinity of other clubs, some with lots of members, some with few. How big is this infinity? Is there any way of matching up clubs and people one to one, thereby showing that the two infinite collections are the same size? Suppose that every person can be paired off with precisely one club, and vice versa. Some of the people will happen to be members of the club with which they are paired (for example, the person who gets paired with the universal club). Others will happen not to be members of their associated club (for example, the person who gets paired with the nul
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In late 2002 and 2003 Grigori Perelman of the Steklov Institute of Mathematics, Saint Petersburg posted 3 papers on the arXiv:[7]\n\n    * The entropy formula for the Ricci flow and its geometric applications, 2002\n    * Ricci flow with surgery on three-manifolds, 2003\n    * Finite extinction time for the solutions to the Ricci flow on certain three-manifolds, 2003\n\nIn these papers he sketched a proof of the Poincaré conjecture and a more general conjecture, Thurston's geometrization conjecture, completing the Ricci flow program outlined earlier by Richard Hamilton.\n\nFrom May to July 2006, several groups presented papers that filled in the details of Perelman's proof of the Poincaré conjecture, as follows:\n\n    * Bruce Kleiner and John Lott, Notes on Perelman's Papers (May 2006) filled in the details of Perelman's proof of the geometrization conjecture.\n    * Zhu Xiping and Cao Huaidong, A Complete Proof of the Poincaré and Geometrization Conjectures - Application of the Hamilton-Perelman theory of the Ricci flow (Asian Journal of Mathematics Volume 10, Number 2, p. 165-492 (June 2006)).\n    * John Morgan and Gang Tian Ricci Flow and the Poincaré Conjecture (July 2006) gave a detailed proof of just the Poincaré Conjecture (which is somewhat easier than the full geometrization conjecture).\n\nOn August 22, 2006, the ICM awarded Perelman the Fields Medal for his work on the conjecture, but Perelman refused the medal.[8][9][10] John Morgan spoke at the ICM on the Poincaré conjecture on August 24, 2006, declaring that "in 2003, Perelman solved the Poincaré Conjecture."[11]\n\nThe August 2006 issue of The New Yorker contains an article, titled "Manifold Destiny", on some of the issues surrounding Perelman's work.
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In mathematics, a symplectic manifold is a smooth manifold M equipped with a closed, nondegenerate, 2-form ω called the symplectic form. The study of symplectic manifolds is called symplectic topology. Symplectic manifolds arise naturally in abstract formulations of classical mechanics and analytical mechanics as the cotangent bundles of manifolds, e.g. in the Hamiltonian formulation of classical mechanics, which provides one of the major motivations for the field: The set of all possible configurations of a system is modelled as a manifold, and this manifold's cotangent bundle describes the phase space of the system.\n\nAny real-valued differentiable function H on a symplectic manifold can serve as an energy function or Hamiltonian. Associated to any Hamiltonian is a Hamiltonian vector field; the integral curves of the Hamiltonian vector field are solutions to the Hamilton-Jacobi equations. The Hamiltonian vector field defines a flow on the symplectic manifold, called a Hamiltonian flow or symplectomorphism. By Liouville's theorem, Hamiltonian flows preserve the volume form on the phase space.
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wow, that's pretty amazing.  i was convinced that was wrong until i tried drawing the "counterexample" i had in my head.  other than reading through that proof (props to the guy who posted that link, btw) i guess the best way to realize this is true is to show that for any five closed shapes in two dimensions, there will always be at least two shapes that do not touch one another.  because of this, you can color those two shapes the same color, so you will have a maximum of four colors.\n\nthis problem is reducible to a graph with five nodes, where each node represents a shape and each edge represents a shared boundary.  (showing that this is reducible to this is probably the difficult part of the proof, which is why i'm glossing over it).  you cannot connect each node to every other node without there being at least one set of intersecting edges.  if edges intersect, it indicates that the shapes (or countries) represented by the two nodes connected by one of the intersecting edges do not share a border.\n\nhere's my suggested proof (assuming we accept the reduction to this graph problem):\nstart with a graph G with three nodes, {A, B, C}.  There are also edges {AB, BC, AC}.  So, G is a closed graph with three nodes and three edges where each node is connected to each other node in the graph (i.e. G is a triangle).  If our reduction of the coloring problem is correct, then there exists a map reducible to this graph that requires three colors to be colored in the described way.\n\nNext, construct a graph H by adding a node D to graph G.  There are two possible ways to add D: 1) D can be on the interior of closed graph G; or 2) D can be outside the interior of G.\n\nIn case 1, let the new graph with nodes A,B,C,D be called H.  We need to show that we can make H such that D is connected to A, B, and C without any intersecting edges.  This is simple enough; straight edges to A, B, and C from D do not intersect each other or the existing edges of the graph G.  So, H demonstrates that a map can require four colors.  H now has four separate enclosed interior regions: triangles ABD, ACD, BCD, and ABC.\n\nFrom H, construct a graph I by adding another node E.  Repeating the above procedure, E can be placed in four places: a) in the interior of ABD; b) in the interior of ACD; c) in the interior of BCD; or d) on the exterior of ABC.\n\nCASE a: As with constructing H, edges can be drawn from E to A, B, and D.  However, C is on the exterior of ABD, so any edge from E to C must cross either AB, BD, or AD.  So, in case a, there must be an intersection, so four colors is still an option.\nCASES b and c follow similarly.\nCASE d: E is on the exterior of ABC, but D is on the interior of ABC.  Any edge between E and D must intersect AB, BC, or AC.  So, coloring with four colors is possible in this case.  Since it is possible in all cases, it is possible for any map with five regions with a sub-map reducible to a graph like H.\n\nCase 2) Instead of placing D in the interior of G, make a graph H' by placing D on the exterior of triangle ABC.  Any quick drawing will show that D is easily connected to A, B, and C.  So there is a map with four regions that requires four colors.  To make the rest of this easy, note that H' also produces four enclosed regions: ABD, ABC, BCD, and ADC.  I won't go through the trouble to prove this, but clearly H' is isomorphic to H.  That being said, whatever is true of the general graph H is also true of the general graph H'.  Therefore, a graph I' made by adding a node to H' cannot be complete without there being at least one set of intersecting edges.\n\nSince this is true for all cases, any map reducible to a graph with five nodes is colorable with just four colors.  QED\n\nSorry this is so long.  I hope it's helpful.
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Euler's Theorem \nA graph divides the plane into several regions called faces. Let us denote the number of the faces by F, the number of the vertices by V, the number of the edges by E. \n the given graph divides the plane into 4 regions, so we have that F=4,V=6,E=8 \n\nProblem . Evaluate F, V and E for the complete graph of 5 vertices, i.e. the pentagon and its diagonals. \n\nAnswer. F=12,V=5,E=10 \n\nA graph that can be drawn in such a way that its edges do not intersect each other (except at their endpoints) is called planar. \nAlthough these two graphs are different, they have equally many vertices and we can numerate the vertices of both graphs in such a way that any two vertices of the first graph are connected by an edge if and only if the corresponding vertices of the second graph (with the same numbers) are connected by an edge. Such graphs are called isomorphic. Actually we will not distinguish isomorphic graphs. So a graph may be planar even if some of its edges intersect in a given picture (see Figure 4). Try to make another couples of pictures describing planar graphs \n\nEuler's Theorem. For a planar graph the equality V-E+F=2 always holds true. \n\n\nRemark. Euler's Theorem is an example of topological obstruction. Problem shows that the requirement for a graph to be a planar is important. An interesting problem is to show that a complete graph is planar if and only if V<5 \n\nProof of Euler's Theorem. For the proof we need to introduce a class of special graphs. A tree G is a connected graph without cycles, i.e. any two vertices of G can be connected by a path (a sequences of edges) and there is no path whose starting and ending vertices coincide. On Figure 5 you can see an example of a tree (a), and a graph which is not a tree (b). \nObviously the trees are simplier then the other graphs. So our goal is to delete some edges until we get a tree, keeping the graph connected. Figure 6 shows how after deleting one edge the quantity V-E+F does nor change. Really, before the operation we have F=5,V=5,E=8 and after that F=4,V=5,E=7. So in both cases V-E+F=2. Since for trees we have that V-E=1 (prove it!) and F=1 the theorem is proved. \nProblem 6. There are 7 lakes in Lakeland. They are connected by 10 canals so that one can swim through the canals from any lake to any other. Haw many islands are there in Lakeland? \n\nHint. Consider the planar graph whose vertices are the lakes, whose edges are the canals, and whose faces are the islands. So F=5. Answer: 4
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zabriskie. merci pour ta reponse a ma question
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This depends on what space the line is embedded in. For example, a line in three dimensions only has one side. So I'm going to assume that you are tlaking about a straight line embedded in the plane. \n\nNow, you have to precisely define what you mean by the phrase 'has two sides'. This is actually non-trivial. We say that a set A embedded in a topological space X has 2 sides if for every open set U containing A, the set U-A is disconnected and there are arbitrarily small such open sets where U-A has exactly two components.\n\nIn the case of a stright line in the plane, we might as well assume that the line, A, is the x-axis. If U is any open set containing A, then U-A={(x,y) in U-A:y>0} union {(x,y) in U-A: y<0}. Both of the sets in this union are open and non-empty, so U-A is disconnected. Furthermore, it is easy to find U so that U-A has exactly these two components. Hence, A is two sided.\n\nFun, but easy, project: Find a space where some embedded line is three sided.
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A fractal is an object or quantity that displays self-similarity, in a somewhat technical sense, on all scales. The object need not exhibit exactly the same structure at all scales, but the same "type" of structures must appear on all scales. A plot of the quantity on a log-log graph versus scale then gives a straight line, whose slope is said to be the fractal dimension. The prototypical example for a fractal is the length of a coastline measured with different length rulers. The shorter the ruler, the longer the length measured, a paradox known as the coastline paradox. \n\nThe term Mandelbrot set is used to refer both to a general class of fractal sets and to a particular instance of such a set. In general, a Mandelbrot set marks the set of points in the complex plane such that the corresponding Julia set is connected and not computable. \n\n"The" Mandelbrot set is the set obtained from the quadratic recurrence equation \n\nZn+1 = (Zn)^2 + C\n\nWhere Z0 = C in the complex plane for all C which this recurrence does not tend toward infinity.
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This is an easy problem.  Here are the rough ingredients:\n\n(1) f is an isometry (i.e. d(f(x),f(y)) = d(x,y) for all x,y in X) implies that f is continuous.  (Prove this!)\n\n(2) The image of a compact set under a continuous map is a compact set, and so f(X) is a compact subset of X.\n\n(3) Suppose, for contradiction, that f is not onto.  So there is x in X &#92; f(X).  Let epsilon be positive real number less than dist(x,f(X)) (this distance is defined to be the infimum of the distances d(x,y) for all y in f(X)).  Since X is compact, X can be cover by finitely many balls of radius epsilon.  Let N be the smallest number of balls of radius epsilon with which you are able to cover X.  x belongs to one of these balls, obviously, and such a ball is disjoint from f(X).  So there is a cover of f(X) consisting of at most N-1 balls of radius epsilon.  But the inverse images (under f) of these balls (all of them of radius epsilon because f is an isometry) cover X.  Since we chose N to be minimal, and we have found now an epsilon-cover of X consisting of N-1 balls, we have reached a contradiction.