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Page "Ringed space" ¶ 5
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If and X
* If numbers have mean X, then.
* If it is required to use a single number X as an estimate for the value of numbers, then the arithmetic mean does this best, in the sense of minimizing the sum of squares ( x < sub > i </ sub >X )< sup > 2 </ sup > of the residuals.
If the method is applied to an infinite sequence ( X < sub > i </ sub >: i ∈ ω ) of nonempty sets, a function is obtained at each finite stage, but there is no stage at which a choice function for the entire family is constructed, and no " limiting " choice function can be constructed, in general, in ZF without the axiom of choice.
If we try to choose an element from each set, then, because X is infinite, our choice procedure will never come to an end, and consequently, we will never be able to produce a choice function for all of X.
If the automorphisms of an object X form a set ( instead of a proper class ), then they form a group under composition of morphisms.
If a detector was placed at a distance of 1 m, the ion flight times would be X and Y ns.
If X and Y are Banach spaces over the same ground field K, the set of all continuous K-linear maps T: X → Y is denoted by B ( X, Y ).
If X is a Banach space and K is the underlying field ( either the real or the complex numbers ), then K is itself a Banach space ( using the absolute value as norm ) and we can define the continuous dual space as X ′ = B ( X, K ), the space of continuous linear maps into K.
* Theorem If X is a normed space, then Xis a Banach space.
If Xis separable, then X is separable.
If F is also surjective, then the Banach space X is called reflexive.
* Corollary If X is a Banach space, then X is reflexive if and only if Xis reflexive, which is the case if and only if its unit ball is compact in the weak topology.
If there is a bounded linear operator from X onto Y, then Y is reflexive.
The tensor product X ⊗ Y from X and Y is a K-vector space Z with a bilinear function T: X × Y → Z which has the following universal property: If T ′: X × Y → Z ′ is any bilinear function into a K-vector space Z ′, then only one linear function f: Z → Z ′ with exists.

If and is
If the circumstances are faced frankly it is not reasonable to expect this to be true.
If his dancers are sometimes made to look as if they might be creatures from Mars, this is consistent with his intention of placing them in the orbit of another world, a world in which they are freed of their pedestrian identities.
If a work is divided into several large segments, a last-minute drawing of random numbers may determine the order of the segments for any particular performance.
If they avoid the use of the pungent, outlawed four-letter word it is because it is taboo ; ;
If Wilhelm Reich is the Moses who has led them out of the Egypt of sexual slavery, Dylan Thomas is the poet who offers them the Dionysian dialectic of justification for their indulgence in liquor, marijuana, sex, and jazz.
If he is the child of nothingness, if he is the predestined victim of an age of atomic wars, then he will consult only his own organic needs and go beyond good and evil.
If it is an honest feeling, then why should she not yield to it??
If he thus achieves a lyrical, dreamlike, drugged intensity, he pays the price for his indulgence by producing work -- Allen Ginsberg's `` Howl '' is a striking example of this tendency -- that is disoriented, Dionysian but without depth and without Apollonian control.
If love reflects the nature of man, as Ortega Y Gasset believes, if the person in love betrays decisively what he is by his behavior in love, then the writers of the beat generation are creating a new literary genre.
If he is good, he may not be legal ; ;
If the man on the sidewalk is surprised at this question, it has served as an exclamation.
If the existent form is to be retained new factors that reinforce it must be introduced into the situation.
If we remove ourselves for a moment from our time and our infatuation with mental disease, isn't there something absurd about a hero in a novel who is defeated by his infantile neurosis??
If many of the characters in contemporary novels appear to be the bloodless relations of characters in a case history it is because the novelist is often forgetful today that those things that we call character manifest themselves in surface behavior, that the ego is still the executive agency of personality, and that all we know of personality must be discerned through the ego.
If he is a traditionalist, he is an eclectic traditionalist.
If our sincerity is granted, and it is granted, the discrepancy can only be explained by the fact that we have come to believe hearsay and legend about ourselves in preference to an understanding gained by earnest self-examination.
If to be innocent is to be helpless, then I had been -- as are we all -- helpless at the start.

If and algebraic
If its minimal polynomial has degree, then the algebraic number is said to be of degree.
If K is a number field, its ring of integers is the subring of algebraic integers in K, and is frequently denoted as O < sub > K </ sub >.
If a is algebraic over K, then K, the set of all polynomials in a with coefficients in K, is not only a ring but a field: an algebraic extension of K which has finite degree over K. In the special case where K = Q is the field of rational numbers, Q is an example of an algebraic number field.
If we allow to incorporate the algebraic sign, this becomes a sum and implies that there is a function of state which is conserved over a complete cycle.
If time, space, and energy are secondary features derived from a substrate below the Planck scale, then Einstein's hypothetical algebraic system might resolve the EPR paradox ( although Bell's theorem would still be valid ).
In 1900, David Hilbert posed an influential question about transcendental numbers, Hilbert's seventh problem: If a is an algebraic number, that is not zero or one, and b is an irrational algebraic number, is a < sup > b </ sup > necessarily transcendental?
If ( a + b ) and ab were both algebraic, then this would be a polynomial with algebraic coefficients.
In universal algebra, a subalgebra of an algebra A is a subset S of A that also has the structure of an algebra of the same type when the algebraic operations are restricted to S. If the axioms of a kind of algebraic structure is described by equational laws, as is typically the case in universal algebra, then the only thing that needs to be checked is that S is closed under the operations.
If ƒ: A → B is a homomorphism between two algebraic structures ( such as homomorphism of groups, or a linear map between vector spaces ), then the relation ≡ defined by
* If α is an algebraic integer then is another algebraic integer.
* If P ( x ) is a primitive polynomial which has integer coefficients but is not monic, and P is irreducible over Q, then none of the roots of P are algebraic integers.
If all axioms defining a class of algebras are identities, then the class of objects is a variety ( not to be confused with algebraic variety in the sense of algebraic geometry ).
If the situation is as above but the extension L of K is algebraic of infinite degree, then it is still possible for the integral closure S of R in L to be a Dedekind domain, but it is not guaranteed.
If P is a common point of two plane algebraic curves X and Y that is a non-singular point of both of them and, moreover, the tangent lines to X and Y at P are distinct then the intersection multiplicity is one.
If a is algebraic over K, then there are many non-zero polynomials g ( x ) with coefficients in K such that g ( a ) = 0.
If the problem of the quadrature of the circle is solved using only compass and straightedge, then an algebraic value of pi would be found, which is impossible.
If X is an affine algebraic set ( irreducible or not ) then the Zariski topology on it is defined simply to be the subspace topology induced by its inclusion into some Equivalently, it can be checked that:
If H is a left module over the ring R, one forms the ( algebraic ) character module H * consisting of all abelian group homomorphisms from H to Q / Z.

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