Search

4 Sep 2002: If you take all invertible affine transformations (that is, transformations of the form x goes to Axb where A is an invertible linear map) then you get affine geometry. ... If you take all homeomorphisms (continuous maps with continuous inverses) then

7 Aug 2002: i) (there exists b> 0) (for all y in X) d(x,y) < b implies d(f(x),f(y)) < a. ... ii)' (there exists b > 0) (for all y in X) d(x,y) < b implies f(y) is in U.

12 Oct 2002: Suppose that A,B : R Rn L arecontinuous and that M ‖A(t,x)‖, ‖B(t,x)‖ for all t and x. ... variable. If suppf B(0,r) then. |f̂(z)| CN(1+‖z‖)Ner|=z|. for all z and integers N 0.

9 Sep 2002: Because (, ) is non-degenerate on i(g), for all B and all sufficiently small,H is a non-degenerate quadratic form on h. ... α,β〉〈α,α〉 for all α,β B is called. the Cartan matrix [13].

4 Oct 2002: i)' (there exists b> 0) (for all y in X) |x-y|< b implies |g(f(x))-g(f(y))|< a. ... iv) (there exists b> 0) (for all y in X) |y-x|< b implies |f(y)-f(x)|< s.

21 Jun 2002: It follows by Definition 1.2 that all invariants ofpositive degree vanish on Z1. ... uχ(h+) = (θV (h+),θW (h+))u for all h+ H5. (24). 20.

10 May 2002: function from B to C then gh consists of all pairs (a,c) in AxC such that there exists b in B with (a,b) in H and (b,c) in ... This would be true if, whenever h(a) took several different values in B, g sent all those values to the same point in C.

7 Aug 2002: ii) (for all b > 0) (for all z) (there exists c> 0) (for all w) |w-z|< c implies |f(w)-f(z)| < b. ... So to deal with this problem first remove "for all b> 0" from the beginning of (ii)' by declaring that b is to be chosen later.

29 Jan 2002: Suppose we know that multiplication and addition satisfy all the axioms for a field. ... Thus, from the existence of multiplicative inverses we can deduce the solutions of all equations of the form ax=b (with a not equal to zero).

7 Aug 2002: Let us write out these statements formally. (ii) (for all u in X) (for all b > 0) (there exists c > 0) (for all v in X) d(u,v) < c implies ... ii)'' (for all b > 0) (there exists c > 0) (for all y in X) d(x,y) < c implies d(f(x),f(y)) < b.

29 Jan 2002: Yes it is: all our numbers are integer combinations of things like 2. ... But what is the dimension of Q(b) over Q(a)? It means the dimension of the vector space of all linear combinations of powers of b, with coefficients in

6 Mar 2002: We could ask ourselves whether we can see how to construct any injection from A to B at all. ... b in B (that is, some non-zero b (mod p)) such that the set of all bx.

25 Nov 2002: Now I shall show, this time by induction on mn, that mn=nm for all positive integers m and n. ... We also know that CxB is a subset of AxB, and that the set-theoretic difference AxB-CxB consists of all ordered pairs (m,x) such that x is in B.

10 Sep 2002: using expressions like w17 if all I do is regard them as notation. ... Then either [a] is minimal or there is a non-empty subset B consisting of all [b] such that b is isomorphic to an initial segment I(t) of a.

12 Sep 2002: be k vertices such that all edges joining them have the same colour. ... constant b < 4 such that R(k) bk for all sufficientlylarge k?

12 Sep 2002: If. you ask whether this mysteriously complicated work has practical applications (and we all. ... was not all that much larger than B. 12. Suppose we now choose a new set of numbers as our set A and that A A is much.