is there an algebraic space with cardinality larger than that of the rationals and smaller than the reals?
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The statement that no such space exists is called the Continuum Hypothesis. It was hypothesized by Cantor, but he gave no proof of it. Godel showed that it is consistent with the usual axioms of set theory to assume CH. Then Paul Cohen showed that it is also consistent with set theory to assume CH is false. In other words, the Continuum Hypothesis is independent of the axioms of set theory. You can either assume it or assume its negation without introducing inconsistencies.\n\nThat said, there are mathematicians that are trying to find other axioms that would either imply CH or its negation. These other axioms would have to have some independent reason for including them in the system, of course. Typically, their attempts have to do with the existence of 'large cardinals'.
710476
Outside the space there is more of Space\nSpace is Infinite
324083
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.
295853
It would be countable and uncountable *sets*, not numbers.\n\nWe say that a set is countable if it is in one-to-one correspondence with some subset of the set of natural numbers (1,2,3,...). The collection of rational numbers turns out to be countable.\n\nWe say that a set is uncountable if it is not countable. ;) The set of real numbers turns out to be uncountbale, but there are many, many sets that are even larger.\n\nWe say that a set A is a subset of the set B if every element of A is in B, but not necessarily the other way around. So the set of natural numbers form a subset of the set of real numbers.\n\nA tensor is a multilinear map on a vector space. Way too much theory for this space.\n\nGiven a *function* f, an inverse function g, is one where the compositions fog and gof are both the identity map.\n\nThe gap between +ve and -ve infinity makes no real sense unless you simple mean the set of real numbers.
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The set of all rational numbers are countable in the following pattern..\nhttp://www.homeschoolmath.net/teaching/rational-numbers-countable.php\nNow from this you can easily verify if you multiply all the numbers by 2/3 then you will get the set you want..So this set is also countable..
79530
http://mathforum.org/library/drmath/sets/high_calculus.html
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There are a few cases that need to be considered here.\n\n1. If n = 0, then n^2 would smaller and 2^n would be larger. (coz 1 is always greater than 0 ;) )\n\n2. If n = 2, 2^n would be smaller and n^2 would be larger.\n\n3. If n < 0 and with decreasing negative values, 2^n would become smaller and smaller and n^2 would become larger and larger.(This case is valid if and only if the instance size can be negative)\n\n4. If n > 26, then n^2 is smaller and 2^n is larger.\n\n5. I haven't considered 'n' as a fraction here coz that will mighty complicate things. \n\nOverall, it was a nice question! Keep posting such questions!
245111
As with many things, this depends on what you mean by 'more'. The collection of even integers is a proper subset of the set of integers, so in that sense there are more integers than even integers. \n\nBut when mathematicians talk about this issue, more or less is described by something called 'cardinality'. We say two sets have the same cardinality if there is a way of pairing off the two sets. In this sense, there are the same number of integers as even integers because you can pair every integers with its double.\n\nIt should be pointed out that there are infinite sets that have different cardinalities. The collection of all real numbers is larger than the set of integers even in this sense.
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Not neccessarily. For instance, the square root of two (which is irrational) multiplied by the square root of two is two, which is very rational.\n\nGood Luck!\n\nP.S. Great Screenname. I love the phantom.
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Using the standard "construction" of the natural numbers, they ARE NOT embedded in the real numbers.  The real numbers are like a completion of a splitting field of a class of polynomials over the natural numbers. (Existance of transcendental numbers, such at pi and e, show that the real numbers themselves are not the splitting field of the set of polynomials over the naturals.  See any good textbook on abstract algebra for definitions of splitting fields, transcendental numbers, ...  )  As such, the natural numbers and real numbers are indeed disjoint sets.  HOWEVER!  Consider the function f which takes 1_N ( 1 in the natural numbers ) to 1_R ( 1 in the real numbers ), 2_N to 2_R, ... etc.  \n\nf( m_N + n_N ) = m_R + n_R\nf( m_N * n_N ) = m_R * n_R\n\nThese properties show that the map f is a homomorphism.  Moreover, the only element which maps to 0_R is 0_N showing that the map is, indeed, an embedding.\n\nWhile the sets themselves may be disjoint, the set of natural numbers is "isomorphic" to what we consider the natural numbers sitting inside the reals.  As we typically consider such sets only "up to isomorphism", we can thus assume that the natural numbers do indeed sit inside the reals.
314860
First of all, aleph_0 is the cardinality of the natural numbers (also, the cardinality of the integers, the rationals, and the algebraics). The cardinality of the real numbers is usually denoted by C (for the continuum). The continuum hypothesis is the conjecture that C=aleph_1. It turns out that this is independent of the usual axioms of set theory. In other words, whether C=aleph_1 or not is undecidable in the sense of Godel.\n\nEven if C=aleph_1, it is not provable that the collection of subsets of R (which is the same as the collection of geometric curves, suitably defined) is the same as aleph_2. So what you ask for is not only not proven, it is not provable. There is another hypothesis called the Generalized Continuum Hypothesis that states that the collection of subsets of a set with cardinality aleph_n has cardinality aleph_{n+1}. This result turns out to be independent of set theory also.
980349
Domain:  all real numbers\nRange:  [2, positive infinity]   (all real numbers from 2 to positive infinity)
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I don't feel like doing the review, but I can offer something.  It turns of that there are functions can are LeBesgue (sp?) integrable but no Riemann.  The f(x) = 0 x rational, =1 x irrational is an example.  You can cook up a set that lands on just the rationals so that it has a limit at a given rational, and hence every rational even though both the rationals and irrationals are infinitely dense in any small neighborhood of any point.  I don't think that you can do the same with irrationals.  If I'm correct then the rationals would be continuous and the irrationals not.  But I don't think that I'm correct.\n\nYou are essentially talking about what's known as point set topology, a part of an undergraduate, or graduate, real analysis course.\n\nI've found Wikipedia to be very helpful for math concepts.\n\nI just found something that said\n\nf(x) = 0, x irrational\nf(x) = x, x rational\n\nis continuous at one point, x = 0.
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This is a Frequently Asked Question. Following is my previous answer:\n\nThis question may be asked so frequently (along with the infamous 0.999...) because no concrete axioms or definitions are ever given for the real number system until one studies higher algebra or real analysis; it is assumed that everyone knows what a real number is until then. The proliferation of these questions show that this assumption is no good.\nFor the sake of conciseness, I will not define what a real number is here, as only the algebraic structure of the real numbers is necessary to answer this question. The algebraic structure of the real numbers is what we call a field. A field is an algebraic structure that follows a finite set of axioms. These axioms define what statements are true and false in this system based on the laws of logic. You can find the field axioms here: http://mathworld.wolfram.com/Field.html .\nEach line is preceded by "For all real numbers a, b and c, the following is true...". Note that the proper form of the additive identity and multiplicative identity is that it prescribes the existence of some number 1 and some number 0 such that the equations hold. These same numbers are then used to prescribe the existence of additive and multiplicative inverses.\nDivision a/b is just shorthand for a*(b^(-1)). So in your question, you are looking for 1*0^(-1). It is easy to show that there is no such number as 0^(-1) in any field. Suppose 0^(-1) exists.\nThen 0*0^(-1) = 1. But multiplying both sides by 0^(-1) gives 0 = 0^(-1), which means 0*0 = 0 = 1. This is obviously false.\nSuppose 0 = 1. 0*a = a for all numbers a. But 0*a = (1 + (-1))*a = a + (-a) = 0. Thus 0 is not equivalent to 1 and our sole assumption that 0^(-1) exists must be false. Thus the expression 1*0^(-1) is undefined in a field.\nProof by contradiction is very powerful for many types of statements.\nAs to the limit argument, note that you are evaluating a limit of the function f(x) = 1/x. As we approach 0 from the negative x-axis, this limit decreases without bound, but as we approach 0 along the positive x-axis, the limit increases without bound. Thus, the limit of f(x) at 0 does not exist, so f(0) remains undefined with no continuous way to complete it. The limit is not indeterminate for this function; it does not exist.\nWhile 1/0 is not defined in any number system, |1/0| is defined as positive infinity in extended number systems (which are not fields by the above argument) for various purposes. See http://mathworld.wolfram.com/AffinelyExtendedRealNumbers.html .
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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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yes, there definately should be because sedonions are just 16-dimensional numbers, if you can have an infinite amount of dimensions... do the math
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I don't remember well this kind of things.\n\nAnyway the key is in the fact that between any two rationals there always exist a non-rational: so there are no intervals of rationals which form a continuum\n\nSorry I don't remember exactly what does the term "open" mean