Simple Abelian Topological Groups. Luke Dominic Bush Hipwood. Mathematics Institute
|
|
- Jerome Taylor
- 5 years ago
- Views:
Transcription
1 M A E NS G I T A T MOLEM UNIVERSITAS WARWICENSIS Simple Abelian Topological Groups by Luke Dominic Bush Hipwood supervised by Dr Dmitriy Rumynin 4th Year Project Submitted to The University of Warwick Mathematics Institute 1 st April, 2014
2 Table of contents 1 Introduction 1 2 Preliminaries 2 3 Locally Compact Case 6 4 Non-complete Example 7 5 Non-metrizable example 17 6 The general Conjecture 24 7 Appendix 26 ii
3 1 Introduction In the following we take the convention 0 / N. All sums are over N unless explicitly stated. We write T to be the subgroup {z C : z = 1} of C, equipped with the subspace topology. For a group G, and an element x G we denote x to be the cyclic subgroup of G generated by x. For a subset Y of a topological space X, denote the closure of Y in X by Y (usually X will be a group and Y a subgroup.) If there exists a map that is both an isomorphism and a homeomorphism between two topological groups G and H, we say that they are topologically isomorphic. In this case we write G = H. If they are isomorphic as groups only, we still write G = H but explicitly say as groups. In 1976 R.C.Hooper gave an example of an infinite, complete, metrizable, topological group whose only locally compact subgroup is the trivial one [Ho]. After which he asked the question whether, if H is an infinite, complete, metrizable, topological abelian group, H has a non-trivial, proper closed subgroup. [Ho] Definition 1.1. An abelian topological group G, is said to be topologically simple if G {e} and contains no closed subgroup, other than the trivial one and itself. Let G be a complete, metrizable, topologically simple, abelian group. Suppose every infinite, complete, metrizable topological abelian group has a non-trivial, proper closed subgroup. It follows that G cannot be infinite. As G is metrizable and finite, it s topology must be the discrete topology. Hence the condition that G contains no closed subgroup, other than the trivial one and itself becomes G contains no non-trivial proper subgroup. That is, G is a simple group. Hence G is topologically isomorphic to C p with the discrete topology, where C p denotes the cyclic group of order p, and p is a prime number. We write G = C p in this case. Conversely if a complete, metrizable, topologically simple abelian group, is necessarily C p with the discrete topology (up to isomorphism), then, certainly, there can not be an infinite, complete, metrizable, topological abelian group that has a no non-trivial, proper closed subgroups. Thus we can reformulate Hooper s conjecture as If G is a complete, metrizable, topologically simple, abelian group, then G is topologically isomorphic with C p equipped with the discrete topology, where p is prime. This project investigates Hooper s conjecture. There are three main chapters. The first proves the conjecture to be true when G is assumed to be locally compact. We in 1
4 fact do this with Hausdorff in place of metrizable, and without the complete assumption altogether. We make use of the heavy machinery available with regards to locally compact abelian groups. In the second we constructed an example of an infinite, metrizable, topologically simple abelian group. This is done by placing an appropriate norm on the integers. We owe thanks to Dr Dmitriy Rumynin, Supervisor, for a sketch to this example. In our final main chapter we give an example of an infinite, complete, topologically simple, Hausdorff abelian group. To do this we construct a net of complete, metrizable abelian groups, each a completion of Z, and such that the subgroup Z is topologically simple. Then taking the inverse limit, we recover Z, with the required topology. We owe thanks to Dr Stanislav Shkarin [S] for a sketch of this example. Our final chapter looks at the general conjecture. We look at the properties that a counter example must have. 2 Preliminaries Here we state a few results that will be used in the proceeding. They are listed in the approximate order in which they are used. Proposition 2.1. [HR, 5.3]. Let G be a topological group, and H a subgroup. Then H is also a subgroup. Proposition 2.2. [HR, 7.1]. Let G be a topological group and C the connected component of the identity. Then C is a closed normal subgroup of G. Proposition 2.3. [HR, 7.7]. Let G be a totally disconnected, locally compact group. Then every neighbourhood of the identity contains a compact open subgroup. Proposition 2.4. [HR, 9.14]. Let G be a locally compact abelian group, and let C be the connected component of the identity. Then C = R n E where n is a non-negative integer and E is a compact connected abelian group. Definition/Proposition 2.5. [HR, ]. Fix a sequence of integers a = (a i ), with a i 2 i. Let a-adic integers be denoted by a. We define the topological group a as follows. As a set a = {0, 1,..., a i 1}. i=1 2
5 For each n N, put Λ n = {x = (x i ) a : x 1 = x 2 = = x n 1 = 0}. {Λ n } n N, forms an open basis of the topology at 0 = (0, 0,... ) (which will be the identity), on a. Given x, y a, x + y is defined as follows. If y = 0 = (0, 0,... ), we set x + 0 = 0 + x = x, and similarly if x = 0. Suppose x, y 0. Let n and m, be the least integers such that x n, y m 0. Without loss of generality, n m. For i < n, set z i = 0. For i = n, write x n + y n = t n a n + z n, with t n Z, and z n {0, 1,..., a n 1}. For i > n, write x i + y i + t i 1 = t i a i + z i. Set z = (z i ). We define x + y = z. The above make a into a Hausdorff, locally compact, compact, abelian group, with a dense subgroup isomorphic to Z. For prime p (or indeed any integer greater or equal to 2), p denotes a, where a = (p, p,... ). Let a = (a i ) be sequence of integers with each a i 2. Let a-adic solenoid be denoted by Σ a. The topological group Σ a is defined as follows. Set u to be the element (1, 0, 0,... ) a. Let B be the subgroup {(n, nu)} n Z R a. Then as a topological group Σ a = (R a )/B. It is a compact connected abelain group. Proposition 2.6. [HR, 25.8]. A compact abelian group G is torsion-free if and only if G = Σ m a p P np p, where P is the set of all prime numbers, m and every n p are arbitrary cardinal numbers, and a = (2, 3, 4,... ). Theorem 2.7 (Baire s Category Theorem). A complete metric space is not a countable union of nowhere dense closed sets. Our final example is of an infinite complete non-metrizable topologically simple abelian group. To do this we need to first know what is meant by a complete, nonmetrizable abelian group. Definition 2.8. [HR, 4.11]. Let G be an abelian group, with identity e. For each open neighbourhood of U of e, set V U = {(x, y) G G : x 1 y U}. 3
6 Define U (G) = {V U : U is a neighbourhood of e}, it is called the uniform structure on G. Note that if G is non-abelian then defined above is the left uniform structure on G, with the right uniform structure on G defined analogously. Proposition 2.9. [B, III 3.1]. U (G) is such that: If V U (G) and W V then W U (G); U (G) is closed under finite intersection; Put = {(x, x) G G}. Then V for all V U (G); If V U (G) then V 1 U (G); For every V U (G) the exists W U (G) such that W 2 U (G). Definition [B, I 6.1, I 7.1, II 3.1]. satisfying A filter on G is a set F of subsets of G If U F and V U then V F; F is closed under finite intersections; / F. A filter F is said to be convergent to x G, if for every neighbourhood of U of x, there is a V F such that V U. A filter F is said to be cauchy if for every V U (G) there is a U F such that U U V. A convergent filter is Cauchy, however in general the converse is not true. [B] II 3.1 Definition [B, II 3.3]. G is complete if every Cauchy filter converges. This coincides with the notion of completeness when G is assumed to be metrizable. Also in our non-metrizable we use inverse limits. Proposition [HR, 6.4]. Let {G α } α A be a collection of topological groups. Put G = G α. Then G is Hausdorff if and only if every G α is Hausdorff. α A 4
7 Proposition [B, II ]. Every closed subgroup of a complete group is complete.. Every product of complete groups is complete. Proposition [HR, 6.14]. Let (A, ) be a partially ordered set, {G α } α A a collection of Hausdorff groups and {f βα : G β G α α, β A, α < β} a collection of continuous homomorphisms. Suppose the above forms an inverse mapping system, that is, such that α < β < γ = f γα = f βα f γβ. Then, the inverse limit is a closed subgroup of the product group G α. Corollary α A [B, II 3.5]. Let (G α, f βα ) be an inverse mapping system. If each G α is Hausdorff and complete then so is the inverse limit. Proposition [mc, 3.4]. Let G be an abelian topological group (written additively), with norm. H G a subgroup. Then the quotient space G/H has norm x + H = inf { x + y }. y H Proposition [mc, 3.25]. Let G be a complete abelian group with norm. Suppose H G is a closed subgroup. Then the quotient group G/H is complete. Proposition [HR, 8.3]. Let G be a Hausdorff group. Then G is metrizable if and only if there is a countable open basis at the identity. In this case the metric can be taken to be left invariant. In the final chapter The General Conjecture we briefly look at dual groups. Definition/Proposition [HR, 23.1, 23.2, 23.15]. A character of a topological group G, is a continuous homomorphism G T. The set of characters is made into an abelian group via point-wise multiplication. This group is called the dual group of G and is denoted G. For every compact set F G and every ε > 0, put U(F, ε) = {χ G : χ(x) 1 < ε x F }. With all the sets U(F, ε) as an open basis at 1 G, G is a topological group. Proposition [AT, 9.5.7]. The dual group of the discrete group Z is topologically isomorphic with T. 5
8 Proposition [HR, 24.2]. Let G be abelian topological group, and such that for every x G \ {e}, there exists a character χ G such that χ(x) 1, then G embeds as a topological subgroup into G in the conical way. 3 Locally Compact Case Theorem 3.1. Let G be a topologically simple, locally compact, Hausdorff, abelian group. Then G = C p where p is a prime number. Proof. Let C denote the connected component to the identity e G, it is a closed subgroup of G, hence either C = {e} or C = G. Case 1: C = {e} In this case G is totally disconnected. Proposition 2.3 then says every neighbourhood of e contains a compact open subgroup. Choose a neighbourhood of e, U G (which exists as we are assuming G is Hausdorff and non-trivial), we find a open, hence closed, subgroup H s.t. e H U G, forcing H = {e}. In particular {e} is open, hence G is discrete. It follows that G must be simple as a group, and hence G = C p for some prime p. Case 2: G = C, i.e. G is a connected abelian group. Since G is locally compact we can apply Proposition 2.4, and deduce G = R n E with E a compact connected abelian group. Because G is topologically simple and a product of non-trivial Hausdorff groups is not topologically simple (see appendix 7.1), either G = R or G = E. Since clearly R is clearly not topologically simple we can rule out the first, and deduce G = E. i.e. G is also compact. Let x be a torsion element. Consider the subgroup x. It has finite order and hence, as G is Hausdorff, it is discrete, and therefore closed. So we must have either x = G or x = {e}. But the first is impossible, as the G would be discrete and so not connected. Thus we must have x = e. i.e. G is torsion-free. Proposition 2.6 then says G = (Σ a ) m p P np p, where P is the set of all primes, m and the n p s are arbitrary cardinal numbers, and 6
9 a = (2, 3, 4,... ). Σ a denotes the a-adic solenoid and p denotes the p-adic integers (as described in the Preliminaries 2.5). p has open, and therefore closed, proper subgroup Λ 2 = {x p : x 1 = 0}. So p is not topologically simple. Since a product of non-trivial Hausdorff groups is not topologically simple, n p = 0 p P. (R Λ 2 )/B is an open and closed subgroup of Σ a, thus Σ a is not topologically simple. Hence m = 0 also. But then G is trivial. Since we excluded G = {e} in the definition of topologically simple, G is not simple; a contradiction. Meaning case 1 is the only possibility. 4 Non-complete Example In this chapter we aim to prove the following theorem: Theorem 4.1. There exists an infinite metrizable, topologically simple, abelian group. To do this we construct a norm on Z. This norm ensures that 1 is in the closure of every subgroup, that is, ensures topological simplicity. Construction. Consider the following sequences: (y n (1) ) : 1 1 1, 2 1 1,..., n 1 1, (n + 1)1 1,... (y n (2) ) : 1 2 1, 2 2 1,..., n 2 1, (n + 1)2 1, (y n (m) ) : 1m 1, 2m 1,..., nm 1, (n + 1)m 1,... (y n (m+1) ) : 1(m + 1) 1, 2(m + 1) 1,..., n(m + 1) 1, (n + 1)(m + 1) 1, Notice for every m N, the sequence (y n (m) ) as n. We define the sequence (x n ) as follows. Set x 1 = 1. Choose x 2 in (y n (1) ) to be such that x 2 > 2 2 x 1. (This is possible since (y n (1) ) ). Now pick x 3 in (y n (2) ) such that x 3 > 2 3 x 2. (Again possible as (y n (2) ) ). Once x k(k+1) 2 is defined (we write K = k(k+1) 2 for ease of notation) choose: x K+1 in (y n (1) ) such that x K+1 > 2 K+1 x K ; x K+2 in (y n (2) ) such that x K+2 > 2 K+2 x K+1 ;... ; x K+k+1 in (y n (k+1) ) such that x K+k+1 > 2 K+k+1 x K+k. As K + k + 1 = (k+1)(k+2) 2, this defines (x n ) inductively. It is such that: 7..
10 (I) x 1 = 1; (II) for every m N the sequence (y n (m) ) contains a subsequence, that itself is a subsequence of (x n ); (III) x n+1 > 2 n+1 x n n N. Now note that for every a Z we can always write a = a i x i for some a i Z all but finitely many 0. This is because x 1 = 1, and therefore ax 1 is a possibility. For a Z we define a = inf a i 2 i where the infimum ranges over all possible ways of writing a = a i x i as above. The sequence (x n ) is the crucial ingredient in proving Theorem 4.1 (and in fact plays a vital role in proving Theorem 5.1 later). Property (I) allows us to define the norm, and will prove useful throughout this and the next chapter. (II) is the key for the simplicity. The purpose of (III) is to ensure we have indeed defined a norm, and will be our main tool in approximating values of. We start by calculating the value of on a certain set of integers. The techniques used in the proof of Proposition 4.2 will be crop up throughout this chapter. Proposition 4.2. Let e = e i x i, with e i {0, 1} all but finitely many zero. Then e = e i 2 i Proof. Let e = x i1 + x i2 + + x in. We proceed by induction on n. Suppose n = 1, then e = x k for some k N. Clearly x k 2 k. Assume x k < 2 k. So we can write x k = a i x i with all but finitely many a i = 0 and a i 2 i < 2 k. We see that a i = 0 i k and that a i < 2 i k i k + 1. Now also for every l N, l a i x i < x l+1. Since otherwise, because x i+1 > 2 i+1 x i i, we would have i=1 l a i x i x l+1 i=1 > (2 l+1 1)x l + x l > (2 l+1 1)x l + (2 l 1)x l 1 + x l 1. > (2 l+1 1)x l + + (2 3 1)x 2 + x 2 > (2 l+1 1)x l + + (2 3 1)x x 1. 8
11 Therefore would require at least one i with a i > 2 i+1 1, contradicting a i < 2 i k i. Let m be maximal with a m 0. By the above m k + 1. So x k = a k+1 x k+1 + a k+2 x k a m x m. ( ) If a m < 0 then as x k > 0, we would require a 1 x a m 1 x m 1 > x m. This l contradicts a i x i < x l+1 l. So a m > 0 (a m 0 by definition). i=1 Now for each i, x i+1 > 2 i+1 x i. This gives a m x m x k x m x k > x m x k+1 > (2 m 1)x m 1 + x m 1 x k+1 > (2 m 1)x m 1 + (2 m 1 1)x m 2 + x m 2 x k+1. > (2 m 1)x m (2 k+3 1)x k+2 + x k+2 x k+1 > (2 m 1)x m (2 k+3 1)x k k+2 x k+1 x k+1 > (2 m 1)x m (2 k+3 1)x k+2 + (2 k+2 1)x k+1 Therefore, for ( ) to hold, we require at least one i {k + 1,..., m} such that a i > 2 i+1 1. Contradicting a i < 2 i k i. Thus x k 2 k, and hence x k = 2 k. Now let n > 1. Clearly e e i 2 i. Assume e < e i 2 i. Write e = a i x i, with all but finitely many a i = 0 and a i 2 i < e i 2 i. As e i 2 i < 1, we see l a i < 2 i i. As in n = 1 case, we also have a i x i < x l+1 l (same proof works here too). Let m be maximal with a m 0. We have e = x i1 + x i2 + + x in = a 1 x 1 + a 2 x a m x m. ( ) Similarly to the n = 1 case, a m > 0. Suppose m > i n. Then n a m x m (x i1 + x i2 + + x in ) > 2 2 x 1 + (2 ij+1 2)x ij + i=1 j=1 m i 1,i 1,...,i n (2 i+1 1)x i Hence for ( ) to hold we need at least one j {1,..., m 1} with a j > 2 j+1 2. This contradicts a i < 2 i i. So m i n. 9
12 If a in = 0 (i.e. m < i n ), then, a 1 x a m x m > x in x m+1 > (2 m+1 1)x m + + (2 3 1)x x 1. Meaning we would need an i with a i > 2 i+1 1; a contradiction. So m = i n. We know from above a in = a m > 0. Consider f = e x i1 = x i1 + + x in 1. By the induction hypothesis, f = 2 i i n 1. But also f = a 1 x 1 + a 2 x (a in 1)x in. Since a a a in 2 in < 2 i in and a in > 0, a a a in 1 2 in < 2 i i n 1. Thus f < 2 i in 1 ; a contradiction. So e = e i 2 i. The result now follows by induction. Now to begin proving we have constructed an infinite, metrizable, topologically simple, abelian group. Lemma 4.3. defines a norm on Z. That is, satisfies (i) a 0 a Z; (ii) a = a a Z; (iii) a = 0 a = 0; (iv) a + b a + b a, b Z. Proof. (i) is obvious. For (ii) we can note that, if a = a i x i then a = a i x i, and hence a = a. (iv): Let a, b Z. Write a = a i x i and b = b i x i, with a i, b i Z all but finitely many zero. Then clearly, a + b = (a i + b i )x i. We then see, a + b a i + b i 2 i a i 2 i + b i 2 i = a i 2 i + b i 2 i. 10
13 Since a i x i and b i x i are arbitrary ways of writing a and b in this form, this inequality holds for every possible way for writing a and b in this form. Hence we indeed have, a + b a + b. (iii): Clearly 0 = 0. Suppose a 0. We need to show a 0. Since a = a, we may assume a > 0. Now as (x i ) and is strictly increasing, there is a (unique) N such that x N a < x N+1. We claim a 2 N. In particular a 0. If a = x N, then, by Proposition 4.2, a = 2 N. So suppose x N < a < x N+1. Assume, for a contradiction, that a < 2 N. Write a = a i x i with a i 2 i < 2 N. We see that a i = 0 i N, and a i < 2 i N i > N. n Now, as in the proof Proposition 4.2, a i x i < x n+1 n (same proof holds here). i=1 We will show inductively that the above in fact implies either a i x i < x N or ai x i x N+1, in both cases, a contradiction. First we note that the first possible non-zero a i is a N+1. Since x N+1 > x N it is clear that either a N+1 x N+1 < x N or a N+1 x N+1 x N+1. Now let k N + 1, and suppose either a N+1 x N a k x k < x N or a N+1 x N a k x k x N+1. Assume first that a N+1 x N a k x k < x N. If a k+1 0, it is obvious that a N+1 x N a k+1 x k+1 < x N. So assume a k+1 > 0. Since each a i < 2 i N, we have a N+1 x N a k+1 x k+1 a N+1 x N a k x k + x k+1 (2 1 1)x N+1 (2 k N 1)x k + x k+1. Now for each i, x i+1 > 2 i+1 x i. This gives x k+1 > (2 k+1 1)x k + + (2 N+3 1)x N N+2 x N+1 > (2 k N 1)x k + + (2 2 1)x N+2 + 2x N+1. Putting these two inequalities together we get, a N+1 x N a k+1 x k+1 > x N+1. Now assume a N+1 x N a k x k x N+1. If a k+1 0, it s obvious that a N+1 x N a k+1 x k+1 x N+1. So suppose a k+1 < 0. Because a N+1 x N a k x k < x k+1, it s clear that a N+1 x N a k+1 x k+1 < 0 < x N. Since the sum a i x i must be a finite sum we indeed have either a i x i < x N or ai x i x N+1. That is either a < x N or a x N+1, contradicting x N < a < x N+1. 11
14 Thus a 2 N, verifying (iii). Before we start talking about topological simplicity we need to make sure we indeed have a topological group. We in fact prove a more general result. Lemma 4.4. Let G be an abelian group (written additively). Let : G R be a norm on G, that is, satisfies (i) a 0 a G; (ii) a = a a G; (iii) a = 0 a = 0; (iv) a + b a + b a, b G. Then G with, is a topological group. Proof. Fix c G, and ε > 0, as a = a a G, if a c < ε, then clearly ( a) ( c) < ε. So a a is continuous. Now (a, b) = a + b defines a norm on G G that coincides with the product topology (see appendix 7.2). Fix (c, d) G G and ε > 0. Then if (a, b) (c, d) < ε, then, by the triangle inequality, a + b (c + d) a c + b d = (a, b) (c, d) < ε. Thus addition is continuous. Finally we check the simplicity of our example. Lemma 4.5. Z with this norm is topologically simple. Proof. Let a Z non-zero. We need to prove a = Z. As 1 generates Z, and the closure of a subgroup is again a subgroup, it suffices to prove 1 a. Since a = a we may assume a > 0. We need to find a sequence, (z n ) say, in a converging to 1, or equivalently, such that (z n 1) converges to 0. Now notice that the sequence (y n (a) ) is the sequence (na 1) n N, and by construction contains a subsequence that is itself a subsequence of (x n ), say (x kn ). We can write this sequence in the form (z n 1) for some sequence (z n ) in a. By construction, for each n, z n 1 = x kn = 2 kn. Hence indeed (z n 1) 0. 12
15 With these Lemmas above we have now proven Theorem 4.1. We now ask if we have stumbled upon a counter example to Hooper s conjuncture, i.e. whether Z with is norm is complete. The answer (unsurprisingly) is no. Proposition 4.6. Z, with the norm, is not complete. Proof. This follow form Baire s Category Theorem. Since single points are certainly closed and nowhere dense, and Z is countable, Z, with any metric, can not be complete. We can realise the incompleteness more explicitly. Proposition 4.7. Consider the sequence (s n ) defined by s n = n x i. It is Cauchy, but has no limit in Z. Proof. It is Cauchy since: for m > n > 0, by Proposition 4.2, s m s n = x n x m = 2 (n+1) m < 2 n. This clearly converges to zero as n, m. We will show that (s n ) does not converge in Z. Let a Z be a candidate limit. By 4.2, lim s n = 2 n = 1, so a 0. Let N N be such that x N 1 a < x N. Let s first suppose a < 0. Fix n > N. We claim that s n a 2 N. Assume for a contradiction that s n a < 2 N. So we can write s n a = a i x i with ai 2 i < 2 N. We immediately see that a i = 0 i N and that a i < 2 i N i > N. k As previously (Proposition 4.2 and Lemma 4.3(iii)), we have that a i x i < x k+1 k (the same proof holds here too). Let m the largest number such that a m 0, so s n a = x x n a = a N+1 x N a m x m. Since a < 0, and, x N is larger then both a and x x N 1 clearly i=1 i=1 x N + + x n < a N+1 x N a m x m < 3x N + x N x n. ( ) Suppose m > n. If a m < 0 then, as s n a > 0, a N+1 x N a m 1 x m 1 > x m, 13
16 k which contradicts a i x i < x k+1 k. So a m > 0 (a m 0 by definition). We have So i=1 a m x m x m > (2 m 1)x m (2 N+2 1)x N N+1 x N. a m x m (3x N +x N+1 + +x n ) > (2 N+1 3)x N + n i=n+1 (2 i+1 2)x i + m 1 i=n+1 (2 i+1 1)x i. Hence for ( ) to be true we would need at least one i {N + 1,..., m 1} with a i > 2 i > 2 i N, which is a contradiction. So a i = 0 i > n. Now if a n 0, because ( ) implies x n < a N+1 x N a n x n, we would need a N+1 x N a n 1 x n 1 > x n ; a contradiction. Suppose a n > 1. Then a n x n 2x n > 2[(2 n 1)x n (2 N+2 1)x N N+1 x N ] > (2 n 1)x n (2 N+2 1)x N N+1 x N + 3x N + x N x n. Hence we have the inequality a n x n (3x N + x N x n ) > (2 n 1)x n (2 N+2 1)x N N+1 x N. So, for ( ) to hold, we require at least one a i 2 i ; a contradiction. Thus a n = 1. ( ) now reads x N + + x n < a N+1 x N a n 1 x n 1 + x n < 3x N + x N x n. Subtracting x n from this we get x N + + x n 1 < a N+1 x N a n 1 x n 1 < 3x N + x N x n 1. The same argument now applies to show a n 1 = 1. After this we repeat over and yield a i = 1 i : N + 2 i n. We eventually get x N + x N+1 < a N+1 x N+1 < 3x N + x N+1. 14
17 It s immediate that a N+1 2. But then a N+1 x N+1 2x N+1 > x N N+1 x N > x N+1 + 3x N ; a contradiction. Now we assume a > 0. Fix n N + 2. If a = x N 1 then, by Proposition 4.2, n s n x N = 2 N + 2 i 2 (N+1), and (s j ) cannot converge to a. So assume a > x N 1. Now because a < x N < x 1 + +x N, and x x N a < x N+1 < 2x N+1, 1=1 x N x n < s n a < 3x N+1 + x N x n. Assume that s n a < 2 (N+1) and write a i x i = s n a with a i 2 i < 2 (N+1). We can conclude similarly to the a < 0 case, that a i = 0 for all i N + 1 and i n + 1, and a i = 1 i: N + 3 i n. Hence reducing to the inequality x N+1 + x N+2 < a N+2 x N+2 < 3x N+1 + x N+2. Similar to before, this is a contradiction. We have proved that for every non-zero a Z. Let N be such that x N 1 a < x N. Then if n N + 2, s n a 2 (N+1). In particular (s k ) cannot converge to a. We have already ruled out s k 0 so (s k ) cannot converge to any a Z. Consider the completion of Z with respect to this norm. Denote it by A. A natural question to ask is whether A remains topologically simple. The answer is no. Proposition 4.8. A has a non-trivial proper closed subgroup. Proof. Let s A be the limit of the Cauchy sequence (s n ) in Proposition 4.7. We prove that the subgroup s is closed in A. Being countable ensures s A. To show s is closed we prove that ε > 0 such that ts ε t Z \ {0}. Then for elements ts t s in s, ts t s = (t t )s ε, proving s is discrete, and hence closed. Since addition and subtraction are continuous, ts n ts, and therefore ts n ts. Thus it is sufficient to prove for n large enough ts n ε. Let t be a non-zero integer. As ts n = ts n n, we may assume t > 0. Choose l N such that 2 l 1 t < 2 l. We claim for n l + 2, ts n
18 Suppose ts n < 1 8 for some n l + 2. Write ts n = a i x i, with a i Z all but finitely many a i = 0 and a i 2 i < 1 8. This immediately forces a i < 2 i 3 i. As before k (Proposition 4.2, Lemma 4.3(iii) and Proposition 4.7), we have that a i x i < x k+1 k. Let m N be maximal such that a m 0, so Now ts l < 2 l (x x l ) < 2 l 2x l < x l+1, ts n = tx tx n = a 1 x a m x m. so ts n < x l+1 + tx l tx n < (1 + 2 l )x l l x l l x n, i.e. a 1 x a m x m < (1 + 2 l )x l l x l l x n. Since, a 1 x a l x l < t(x x l ) (as a i < 2 i 3 < 2 l 1 < t i l), and in turn t(x x l ) < 2 l (x x l ) < x l+1, it follows that a l+1 x l a m x m < 2x l+1 + n i=l+1 i=1 2 l x i. (1) Now suppose m > n. If a m < 0, then as ts n > 0, we would need m 1 a i x i > x m ; a contradiction. So a m > 0 (a m 0 by definition). Now therefore a m x m [2x l+1 + a m x m x m > (2 m 1)x m (2 l+3 1)x l l+2 x l+1, n i=l+1 i=1 2 l x i ] > (2 l+2 2 l 4)x l+1 + n (2 i+1 2 l 1)x i + i=l+2 m 1 i=n+1 (2 i+1 1)x i. Thus, for (1) to hold, i {1,..., n 1} with at least a i > 2 i+2 2 l 2. But for all i N, 2 i+2 2 l 2 > 2 i 3. Hence we have a contradiction. So we conclude m n, i.e. a k = 0 k > n. Hence we have We now aim to show a n = t. a 1 x a n x n = tx tx n. (2) If a n < t, then a 1 x a n 1 x n 1 > (t a n )x n x n ; a contradiction. So a n t. 16
19 Assume a n > t, then, because ts n 1 < 2 l (x x n 1 ) < 2 l+1 x n 1 2 (n 1) x n 1 = 1 2 2n x n 1 < 1 2 x n (recall n l + 2), we have a n x n ts n x n ts n 1 > 1 2 x n > 1 2 [(2n 1)x n (2 3 1)x x 1 ]. Thus for (2) to be true we need at least one i {1,..., n 1} with a i > 1 2 (2i+1 1) > 2 i 3 ; a contradiction. Hence indeed a n = t. Equation (2) now reduces to a 1 x a n 1 x n 1 = tx tx n 1. We can now repeat the above over, yielding a l+2 = = a n = t. In particular, a l+2 = t, and thus ai 2 i t2 (l+2) 2 l 1 2 (l+2) = 1 8. Contradicting our initial assumption. So this assumption must be false, that is, ts n 1 8. Hence we conclude s is a proper non-trival closed subgroup, that is, A is not topologically simple. Notice that this proof allows us to see more explicitly that s A. Any non-zero a A with a < 1 8 will not be in s, a = x 4 for example. 5 Non-metrizable example Our aim in this Chapter is to prove Theorem 5.1. There exists an infinite, complete, topologically simple, Hausdorff abelian group. The example we give will turn out to be Z with an non-metrizable topology. This will therefore not be a counter-example to Hooper s conjecture. Our first aim will be to construct a collection complete of groups, each with Z as a subgroup. Furthermore the subgroup Z should be topologically simple in each case. Let A be the set of real sequences a = (a n ) satisfying: 17
20 a n (0, 1] n N; a is non-increasing. a n 0 as n 2 n a n n N. For a, b A write a b whenever a n b n n. This makes A into a partially ordered set. Notice also that A is uncountable. Indeed we have an obvious injection (0, 1] A, r (r, 0, 0,... ). For each a A, we will construct a complete group, G a, with the required properties. Thus A will be our indexing set for these groups. The reason for partially ordering A will become clear later, and is crucial in recovering the topologically simple Z. Fix a A. Consider the subgroup H a of Z ℵ 0, consisting of the sequences h = (h n ) satisfying a n h n <. It is a subgroup, since, if h, h H a, with h = (h n ) and h = (h n), then hn h n a n h n a n + h n a n < by the triangle inequality. So h h H a. For h H a define h a = a n h n. Lemma 5.2. a is a norm on H a and with this norm H a is a topological group. Proof. We need to prove (i) h a 0 h H a ; (ii) h a = h a h H a ; (iii) h a = 0 h = 0 = (0, 0,... ); (iv) h + h a h a + h a h, h H a ; (v) Addition and subtraction are continuous. (i) is obvious. Given h H a h a = a n h n = a n h n = h a 18
21 so (ii) is true. Clearly 0 a = 0. Conversely, if h a = 0, i.e. an h n = 0, then as a n 0 n, h n = 0 n. So h = 0, proving (iii). Given h, h H a, by the standard triangle inequality h + h a = a n h n + h n a n h n + a n h n = h a + h a, so (iv) holds. (v) follows form lemma 4.4. Lemma 5.3. H a, with this norm, is complete. Proof. Let (h (m) ) be a cauchy sequence in H a. This means i a i h (m) i For each i N, since a i > 0, h (m) i h (k) i 0 as m, k. h (k) i 0 as m, k. Therefore every h (m) i Z m, we see h i Z and that h (m) i = h i converges, say to h i. Recalling that h (m) i eventually. We need to show the sequence h = (h n ) is in H a, and that h (m) h. Let ε > 0, and M N be such if m, k > M then i a i h (m) i h (k) i < ε. Indeed take m, k > M. Since each term in the above sum is non-negative, for any n N we certainly have Letting k we get n i=1 n i=1 a i h (m) i h (k) i < ε. a i h (m) i h i ε. Since this holds for all n N, we can send n, yielding i=1 a i h (m) i h i ε, i.e. h (m) h a ε. Hence indeed h (m) h. Note also that the above implies for any m > M, h (m) h H a. As H a is a subgroup of Z ℵ 0, it follows h H a. We now turn our attention to the simplicity of our example. Let (x n ) be the sequence of positive integers constructed in chapter 4. Recall it satisfies: 19
22 (I) x 1 = 1; (II) for every m N the sequence (mn 1) n N contains a subsequence, that itself is a subsequence of (x n ); (III) x n+1 > 2 n+1 x n n N. Note the condition (III) implies x n+1 > 2 1 a n x n n. We make use of the latter property of (x n ), and is the reason we imposed the last condition on A. Let K Z ℵ 0 consist of sequences c = (c n ) such that all but finitely many c n = 0, and c n x n = 0. Clearly K is a subgroup and is contained in H a. Let G a = H a /K. Lemma 5.4. K is closed in H a, and hence the quotient group G a is complete Proof. Claim: c a 1 c K \ {0}. Let c K \ {0}. Suppose c n1, c n2,..., c nk k 2). Then, by definition, c n1 x n1 + + c nk x nk = 0. are exactly all the non-zero terms (note Equivalently c n1 x n1 + + c nk 1 x nk 1 = c nk x nk. As x n+1 > 2 1 a n x n, we therefore have c n1 x n1 + + c nk 1 x nk 1 = c nk x nk x nk > 2 1 a nk 1 x nk 1 > 1 a nk 1 x nk a nk 2 x nk 2. > 1 a nk 1 x nk a n2 x n a n1 x n1. For this to be true we need at least one i with c ni > 1 a ni, say c nl > 1 a nl. But then proving the claim. c a = c i a i > c nl a nl > 1 a nl a nl = 1, The claim shows K is a discrete subgroup, hence it is closed. As a quotient of a complete space by a closed space, G a is complete. We claim that the G a s are the complete groups we are after. Lemma 5.5. Z embeds as a cyclic dense subgroup of G a, via ι : t (t, 0, 0,... ) + K. 20
23 Proof. It is clear that ι is a homomorphism. It is injective since if t ker(ι), then (t, 0, 0,... ) K. So by definition tx 1 = 0. But x 1 = 1, so t = 0. Put T = {(t, 0, 0,... ) : t Z}. For t Z, denote t = (t, 0, 0,... ). We prove T + K is dense in H a, and thus im(ι) is dense in G a. Let h H a, and ε > 0. Since h i a i <, n N such that h i a i < ε. i=n+1 Define c = (c i ) as follows: Put c i = h i for i {2, 3,..., n}, put c 1 = n c i x i and c i = 0 otherwise. Then, by construction, c i x i = 0 (recall x 1 = 1), and hence c K. Set t = h 1 c 1. Then h (t + c) a = h 1 (t + c 1 ) a 1 + as required. i=2 n h i c i a i + h i a i = h i a i < ε i=2 i=n+1 i=n+1 Define Z = {(t, 0, 0,... ) + K : t Z} = im(ι). Notice Z G b b A. Remark. Notice that the above proof shows that Z is the quotient of the subgroup {(h n ) H a : all but finitely many h n = 0} by K. Proof. Let S = {(h n ) H a : all but finitely many h n = 0}, and T be as above. It suffices to prove T + K = S + K. Clearly T S, so T + K S + K. Conversely let s S. Define c in the same way as above, but with s in place of h. Then s c T. Thus S + K T + K. At this point we indeed have our a collection of complete groups, {G a } a A. Each G a has a copy Z sitting inside, namely Z, and 1 is in the closure of every subgroup as required (see Lemma 5.8 for a proof of this). We look to recover Z form these groups. To this we take the inverse limit (which is why we made A into a partially ordered set). Let a, b A with a b. If h H b, h = (h n ), then by definition, h i b i <. since a b, it follows that h i a i <, and hence h H a. So H b H a. Let f ba : G b G a, h + K h + K. Lemma 5.6. For every a, b A with a b, f ba is an injective continuous homomorphism. If a, b, c A, with a b c, then f ba f cb = f ca. 21
24 Proof. Let a, b A with a b. Because K is independent of a and b, and is contained in every Hã, ã A, it is clear that f ba is a well-defined injective homomorphism. It is continuous because h a h b h G b. The last part is obvious. By the above lemma we have an inverse mapping system consisting of A, the G a s and the f ba s. Let Z G a denote the inverse limit. As each G a is metrizable (so a A certainly Hausdroff), Z is a closed subgroup of the complete group G a. Hence Z is complete. We claim Z is the topological group we want. Lemma 5.7. Z = Z as groups. a A Proof. We prove Z = {(t + K) a A a A G a : t + K Z}, which is clearly isomorphic to Z, which in turn is isomorphic to Z by Lemma 5.5. For now, to ease notation, let R denote the right hand side. As Z G a a A, given t + K Z and a, b A with a b, it s clear f ba (t + K) = t + K. Thus R Z. Conversely, assume Z R. Let (h a + K) a A Z \ R. Suppose first that h a Z a A. By choosing different co-set representatives (if necessary) we may assume h a = t a = (t a, 0, 0,... ) a A. Let a, b A (not necessarily comparable). Define c = (c n ), by c n = max{a n, b n }, then c A, and a, b c. f ca (t c + K) = t a + K, i.e. Therefore we must have t c + K = t a + K, and f cb (t c + K) = t b + K, i.e. t c + K = t b + K. Hence t a + K = t b + K. So a, b A, t a + K = t b + K, contradicting (h a + K) a A / R. Hence a A such that h a + K / Z. Write h a = (h i ) i N ; by the remark after Lemma 5.5, there must be infinitely many i such that h i 0. Let (h ni ) be a subsequence of (h i ) consisting entirely of non-zero terms. Define b = (b j ) as follows. For j such that 1 j n 1, put b j = 1; for i 2 and for j such that n i 1 < j n i, put b j = max{a ni 1, 1 i }. Clearly b A, and, by construction, a b (recall ã is non-increasing ã A). So h b + K G b such that f ba (h b + K) = h a + K, i.e. h b + K = h a + K. Hence h a = h b + k, for some k K. In particular h a H b. But a contradiction. Hence Z R. So Z = R, and hence Z = Z. hi b i h ni b ni b ni 1 i = ; 22
25 The topology on Z is the subspace topology from G a. Thus, considering Z with this topology, a sequence (t n ) in Z converges to a limit, say t, if and only if the sequence ((t n, 0, 0,... ) + K) converges to (t, 0, 0,... ) + K in G a, for every a A. a A Lemma 5.8. Z, with the above topology, is topologically simple. Proof. Let t Z, we need to show 1 t. By construction of the sequence (x n ), it has a subsequence, (x kn ) say, such that, for each n N, x kn = ty n 1, for some positive integers y n. Writing z n = ty n, (z n ) is a sequence in t and is such that (z n 1) = (x kn ). Fix a A and n N. Define c 1 = x kn, c kn = 1, c i = 0 otherwise, and put c = (c i ). Then ci x i = x kn + x kn = 0, so c K. Furthermore (z n 1, 0, 0,... ) + c = (0,..., 0, 1, 0,... ), where the 1 is in the k n th position. Hence (z n 1, 0, 0,... ) + K = (0,..., 0, 1, 0,... ) + K. As (0,..., 0, 1, 0,... ) a = a kn 0 as n, it follows that (z n, 0, 0,... ) + K (1, 0, 0,... ) + K in G a. Since this holds a A, z n 1 in Z. Hence 1 t, so t = Z and Z is topologically simple. We have now proven Theorem 5.1, that is we have successfully constructed an infinite, complete, topologically simple, Hausdorff abelian group. Intuitively, we have no hope of Z, with this topology being metrizable; there are just too many norms on it. Indeed, it cannot be since we know there are no countable complete metric spaces by Baire s Category Theorem. A more direct proof is possible. Proposition 5.9. Z is not metrizable. Proof. By proposition 2.18 it suffices to show that there is no countable open basis at 0 = (0 + K, 0 + K,... ). i write Let U = {U i } i N, be a countable collection of open sets, each containing 0. For each U i = {(t + K) a A Z : t + K U a (i) }, with all but finitely many U a (i) = Z (recall Z = {(t + K) a A G a a A : t + K Z}.) As A is uncountable and there are only countably many U i, there are uncountably many a A such that U (i) a = Z i N, 23
26 in particular there is at least one, say b. Let U b be any open subset of G b containing 0 + K, and such that Z U. Then the open set {(t + K) a A Z : t + K U b }, contains 0 but cannot be written as a union of sets form U. Hence there can be no countable open basis a 0, and hence Z is not metrizable. 6 The general Conjecture Suppose G is a counter-example to Hooper s conjecture, that is, an infinite, complete, metrizable, topologically simple abelian group. Let e G denote the identity. We now list the obvious properties that G must have. By Baire s Category Theorem, a complete metric space cannot be countable. Hence G is uncountable. In the locally compact case we proved either G is connected or totally disconnected. The same proof holds in the general case, and hence the same conclusion holds too. Let x G be a torsion element and consider the subgroup x. As x has finite order, x is finite. Since G is metrizable, it is closed. Because G is infinite x G, and so x = {e}, i.e. x = e. Hence G is torsion-free. Pick any x G \ {e}, and consider the subgroup x. It is a non-trivial closed subgroup, hence by assumption, x = G. Now x has infinite order, and therefore as a group x = Z. Because G is complete, the completion of x is just x = G. Thus we can view G as a completion of Z. So G has the following properties: G is uncountable; Either G is connected or G is totally disconnected; G is torsion-free; G is the completion of Z with an appropriate metric. Write G as the completion of Z with respect to some metric. We now consider the dual groups of G and the topological subgroup Z, and see what they can tell us about G. Let Z d denote the group Z equipped with the discrete topology. Proposition 2.20 says Z d = T. A topological isomorphism is given by χ χ(1). Since any continuous map Z T, is certainly a continuous map Z d T, we can embed the group Z into T via χ χ(1). Thus we can view the group Z as a subgroup of T. The question remains whether the topologies necessarily coincides. It is at least true that the topology on Z is finer then the subgroup to topology inherited form T. 24
27 Proof. It suffices to check the topology is finer at 1 T. Let ε > 0. Write B ε (1) = {z C Z : z 1 < 1}. Choose F = {1} Z. Clearly F is compact. By definition (definition/proposition 2.19) U(F, ε) = {χ Z : χ(1) 1 < ε} is an open set of Z. Recalling the natural embedding Z T, is given by χ χ(1), we see U(F, ε) is just B ε (1). Thus topology on Z is indeed finer then that of T. Given a continuous χ : Z T, we can define a map ˆχ : G T as follows. For x Z, define ˆχ(x) = χ(x). For x G \ Z, choose a sequence in Z, say (x n ), converging to x (which exists as Z is dense in G). Define ˆχ(x) = lim χ(x n ). Since χ is continuous, and T is complete, ˆχ is a well-defined continuous homomorphism G T, i.e. ˆχ G. Conversely given a character of G, we can restrict to a character of Z. This procedure clearly gives a topological isomorphism, thus Z = G. Fix a non-zero n Z, and consider the group homomorphism φ : Z T; 1 ζ, where ζ is a primitive nth root of unity. Suppose φ is continuous and let ˆφ be it s unique extension to G. Then ker ˆφ is a closed subgroup of G. ker ˆφ {0} since n ker ˆφ, and 1 / ker ˆφ, so ker ˆφ is a proper non-trivial closed subgroup of G. This contradicts the simplicity of G. Thus φ is not continuous. In particular the kernel, n is not closed, and so the subgroup Z is topologically simple as well. Also notice this shows Z is proper subgroup of T. We now know the following: Z = G ; The subgroup Z is topologically simple also; Z is proper subgroup of T; The topology on Z is at worst finer then T. Now either Z = G = {1}, or Z = G {1}. In the latter case, as G is topologically simple, any character will have trivial kernel, and thus G will satisfy the hypotheses of proposition Hence G conically embeds into G which equals Z. Thus there are two cases Either G = {1}, or G is a topological subgroup of Z. Recalling Z is a proper subgroup of T with some finer topology, call it H T, we have Either G = {1}, 25
28 or G is a topological subgroup of H, where H is a proper subgroup of T with some finer topology. We have successfully managed to reduce the group structures we are working with to the well know Z and T. 7 Appendix Proposition 7.1. Suppose {G α } α A is a collection of non-trivial Hausdorff abelian groups, and G = G α. If G is topologically simple then A = 1. α A Proof. Since we exclude the trivial group in the definition of topologically simple, A 0. Suppose A > 1, possible infinite. Let β A. Set H = {(g α ) G : g α = 0 α β}. Clearly H is a proper subgroup of G. For every α, G α is Hausdorff and so {0} is closed in G α. It follows G α \{0} is open, for every α. For α β put U α = G α \{0}. Set U β =. Then U = {(g α ) : g α U α } is open. Hence H = G \ U is a proper closed subgroup, i.e. G is not topologically simple. Thus A = 1. Proposition 7.2. Let be a norm on an abelian group G, i.e. satisfies (i) - (iv) in lemma 4.4. Define (a, b) = a + b on G G. Then is a norm on G G, compatible with the product topology induced by. Proof. The fact that is a norm on G G is well known and follows readily form being a norm on G. Let B ε (c) = {a G : a c < ε}, and B δ (c, d) = {(a, b) G G : (a, b) (c, d) < δ}. A basis for the topology on G is Therefore the product topology has basis The topology induced by has basis B 0 = {B ε (a) : a G, ε > 0}. B = {U V : U, V B 0 }. B = {B δ (a, b) : (a, b) G G δ > 0}. Given an element of B we need to find an element of B contained in it, and vice versa. 26
29 Take U V B, and assume U = B ε (c) and V = B δ (d). With out loss of generality assume ε < δ. Suppose (a, b) B ε(c, d) then a c a c + b d = (a, b) (c, d) < ε, and similarly b d (a, b) (c, d) < ε δ, so (a, b) U V. Hence B ε(c, d) U V. Conversely, let U B, and assume U = B 2ε (c, d). If a B ε(c) and b B ε (d), then (a, b) (c, d) = a c + b d < ε + ε = 2ε. So B ε (c) B ε (d) U. 27
30 References [Ho] R.C.Hooper, Locally Compact Subgroups of Metrizable Topological Abelian Groups, Proc. Amer. Math. Soc. 57 (1976), no. 1, [HR] E.Hewitt, K.A.Ross, Abstract Harmonic Analysis, Volume 1, Springer-Verlag, Berlin, Göttingen, Heidelberg, 1963 [B] N.Bourbaki, Elements of Mathematics, General Topology, Springer-Verlag Berlin Heidelberg New York London Paris Tokyo 1989 [AT] A. Arhangel skii and M.Tkachenko Topological Groups and Related Structures, Atlantis Press, Amsterdam Paris, World Scientific, 2008 [mc] B.MacCluer, Elementary Functional Analysis, Springer, New York, 2009 [S] Dr Stanislav Shkarin, Private Communication 28
Chapter 2 Metric Spaces
Chapter 2 Metric Spaces The purpose of this chapter is to present a summary of some basic properties of metric and topological spaces that play an important role in the main body of the book. 2.1 Metrics
More informationDefinitions. Notations. Injective, Surjective and Bijective. Divides. Cartesian Product. Relations. Equivalence Relations
Page 1 Definitions Tuesday, May 8, 2018 12:23 AM Notations " " means "equals, by definition" the set of all real numbers the set of integers Denote a function from a set to a set by Denote the image of
More informationMath 426 Homework 4 Due 3 November 2017
Math 46 Homework 4 Due 3 November 017 1. Given a metric space X,d) and two subsets A,B, we define the distance between them, dista,b), as the infimum inf a A, b B da,b). a) Prove that if A is compact and
More information7 Complete metric spaces and function spaces
7 Complete metric spaces and function spaces 7.1 Completeness Let (X, d) be a metric space. Definition 7.1. A sequence (x n ) n N in X is a Cauchy sequence if for any ɛ > 0, there is N N such that n, m
More informationCompleteness and quasi-completeness. 1. Products, limits, coproducts, colimits
(April 24, 2014) Completeness and quasi-completeness Paul Garrett garrett@math.umn.edu http://www.math.umn.edu/ garrett/ [This document is http://www.math.umn.edu/ garrett/m/fun/notes 2012-13/07d quasi-completeness.pdf]
More informationHilbert spaces. 1. Cauchy-Schwarz-Bunyakowsky inequality
(October 29, 2016) Hilbert spaces Paul Garrett garrett@math.umn.edu http://www.math.umn.edu/ garrett/ [This document is http://www.math.umn.edu/ garrett/m/fun/notes 2016-17/03 hsp.pdf] Hilbert spaces are
More information8 Complete fields and valuation rings
18.785 Number theory I Fall 2017 Lecture #8 10/02/2017 8 Complete fields and valuation rings In order to make further progress in our investigation of finite extensions L/K of the fraction field K of a
More informationProblem Set 2: Solutions Math 201A: Fall 2016
Problem Set 2: s Math 201A: Fall 2016 Problem 1. (a) Prove that a closed subset of a complete metric space is complete. (b) Prove that a closed subset of a compact metric space is compact. (c) Prove that
More informationChapter 8. P-adic numbers. 8.1 Absolute values
Chapter 8 P-adic numbers Literature: N. Koblitz, p-adic Numbers, p-adic Analysis, and Zeta-Functions, 2nd edition, Graduate Texts in Mathematics 58, Springer Verlag 1984, corrected 2nd printing 1996, Chap.
More informationPart III. 10 Topological Space Basics. Topological Spaces
Part III 10 Topological Space Basics Topological Spaces Using the metric space results above as motivation we will axiomatize the notion of being an open set to more general settings. Definition 10.1.
More information3 COUNTABILITY AND CONNECTEDNESS AXIOMS
3 COUNTABILITY AND CONNECTEDNESS AXIOMS Definition 3.1 Let X be a topological space. A subset D of X is dense in X iff D = X. X is separable iff it contains a countable dense subset. X satisfies the first
More informationNOTES IN COMMUTATIVE ALGEBRA: PART 2
NOTES IN COMMUTATIVE ALGEBRA: PART 2 KELLER VANDEBOGERT 1. Completion of a Ring/Module Here we shall consider two seemingly different constructions for the completion of a module and show that indeed they
More informationINVERSE LIMITS AND PROFINITE GROUPS
INVERSE LIMITS AND PROFINITE GROUPS BRIAN OSSERMAN We discuss the inverse limit construction, and consider the special case of inverse limits of finite groups, which should best be considered as topological
More informationCHODOUNSKY, DAVID, M.A. Relative Topological Properties. (2006) Directed by Dr. Jerry Vaughan. 48pp.
CHODOUNSKY, DAVID, M.A. Relative Topological Properties. (2006) Directed by Dr. Jerry Vaughan. 48pp. In this thesis we study the concepts of relative topological properties and give some basic facts and
More information1 Topology Definition of a topology Basis (Base) of a topology The subspace topology & the product topology on X Y 3
Index Page 1 Topology 2 1.1 Definition of a topology 2 1.2 Basis (Base) of a topology 2 1.3 The subspace topology & the product topology on X Y 3 1.4 Basic topology concepts: limit points, closed sets,
More informationFILTERED RINGS AND MODULES. GRADINGS AND COMPLETIONS.
FILTERED RINGS AND MODULES. GRADINGS AND COMPLETIONS. Let A be a ring, for simplicity assumed commutative. A filtering, or filtration, of an A module M means a descending sequence of submodules M = M 0
More informationThe small ball property in Banach spaces (quantitative results)
The small ball property in Banach spaces (quantitative results) Ehrhard Behrends Abstract A metric space (M, d) is said to have the small ball property (sbp) if for every ε 0 > 0 there exists a sequence
More informationTopological vectorspaces
(July 25, 2011) Topological vectorspaces Paul Garrett garrett@math.umn.edu http://www.math.umn.edu/ garrett/ Natural non-fréchet spaces Topological vector spaces Quotients and linear maps More topological
More informationTopological properties
CHAPTER 4 Topological properties 1. Connectedness Definitions and examples Basic properties Connected components Connected versus path connected, again 2. Compactness Definition and first examples Topological
More informationVARIETIES OF ABELIAN TOPOLOGICAL GROUPS AND SCATTERED SPACES
Bull. Austral. Math. Soc. 78 (2008), 487 495 doi:10.1017/s0004972708000877 VARIETIES OF ABELIAN TOPOLOGICAL GROUPS AND SCATTERED SPACES CAROLYN E. MCPHAIL and SIDNEY A. MORRIS (Received 3 March 2008) Abstract
More informationTopology Proceedings. COPYRIGHT c by Topology Proceedings. All rights reserved.
Topology Proceedings Web: http://topology.auburn.edu/tp/ Mail: Topology Proceedings Department of Mathematics & Statistics Auburn University, Alabama 36849, USA E-mail: topolog@auburn.edu ISSN: 0146-4124
More informationB 1 = {B(x, r) x = (x 1, x 2 ) H, 0 < r < x 2 }. (a) Show that B = B 1 B 2 is a basis for a topology on X.
Math 6342/7350: Topology and Geometry Sample Preliminary Exam Questions 1. For each of the following topological spaces X i, determine whether X i and X i X i are homeomorphic. (a) X 1 = [0, 1] (b) X 2
More informationMATH 54 - TOPOLOGY SUMMER 2015 FINAL EXAMINATION. Problem 1
MATH 54 - TOPOLOGY SUMMER 2015 FINAL EXAMINATION ELEMENTS OF SOLUTION Problem 1 1. Let X be a Hausdorff space and K 1, K 2 disjoint compact subsets of X. Prove that there exist disjoint open sets U 1 and
More informationStanford Mathematics Department Math 205A Lecture Supplement #4 Borel Regular & Radon Measures
2 1 Borel Regular Measures We now state and prove an important regularity property of Borel regular outer measures: Stanford Mathematics Department Math 205A Lecture Supplement #4 Borel Regular & Radon
More informationDENSELY k-separable COMPACTA ARE DENSELY SEPARABLE
DENSELY k-separable COMPACTA ARE DENSELY SEPARABLE ALAN DOW AND ISTVÁN JUHÁSZ Abstract. A space has σ-compact tightness if the closures of σ-compact subsets determines the topology. We consider a dense
More informationMETRIC HEIGHTS ON AN ABELIAN GROUP
ROCKY MOUNTAIN JOURNAL OF MATHEMATICS Volume 44, Number 6, 2014 METRIC HEIGHTS ON AN ABELIAN GROUP CHARLES L. SAMUELS ABSTRACT. Suppose mα) denotes the Mahler measure of the non-zero algebraic number α.
More informationOverview of normed linear spaces
20 Chapter 2 Overview of normed linear spaces Starting from this chapter, we begin examining linear spaces with at least one extra structure (topology or geometry). We assume linearity; this is a natural
More informationA Version of the Grothendieck Conjecture for p-adic Local Fields
A Version of the Grothendieck Conjecture for p-adic Local Fields by Shinichi MOCHIZUKI* Section 0: Introduction The purpose of this paper is to prove an absolute version of the Grothendieck Conjecture
More informationMaths 212: Homework Solutions
Maths 212: Homework Solutions 1. The definition of A ensures that x π for all x A, so π is an upper bound of A. To show it is the least upper bound, suppose x < π and consider two cases. If x < 1, then
More informationWinter School on Galois Theory Luxembourg, February INTRODUCTION TO PROFINITE GROUPS Luis Ribes Carleton University, Ottawa, Canada
Winter School on alois Theory Luxembourg, 15-24 February 2012 INTRODUCTION TO PROFINITE ROUPS Luis Ribes Carleton University, Ottawa, Canada LECTURE 2 2.1 ENERATORS OF A PROFINITE ROUP 2.2 FREE PRO-C ROUPS
More informationMeasurable Choice Functions
(January 19, 2013) Measurable Choice Functions Paul Garrett garrett@math.umn.edu http://www.math.umn.edu/ garrett/ [This document is http://www.math.umn.edu/ garrett/m/fun/choice functions.pdf] This note
More informationB. Appendix B. Topological vector spaces
B.1 B. Appendix B. Topological vector spaces B.1. Fréchet spaces. In this appendix we go through the definition of Fréchet spaces and their inductive limits, such as they are used for definitions of function
More informationMath 676. A compactness theorem for the idele group. and by the product formula it lies in the kernel (A K )1 of the continuous idelic norm
Math 676. A compactness theorem for the idele group 1. Introduction Let K be a global field, so K is naturally a discrete subgroup of the idele group A K and by the product formula it lies in the kernel
More informationON MATCHINGS IN GROUPS
ON MATCHINGS IN GROUPS JOZSEF LOSONCZY Abstract. A matching property conceived for lattices is examined in the context of an arbitrary abelian group. The Dyson e-transform and the Cauchy Davenport inequality
More informationA NICE PROOF OF FARKAS LEMMA
A NICE PROOF OF FARKAS LEMMA DANIEL VICTOR TAUSK Abstract. The goal of this short note is to present a nice proof of Farkas Lemma which states that if C is the convex cone spanned by a finite set and if
More informationMath 210B. Artin Rees and completions
Math 210B. Artin Rees and completions 1. Definitions and an example Let A be a ring, I an ideal, and M an A-module. In class we defined the I-adic completion of M to be M = lim M/I n M. We will soon show
More informationGALOIS EXTENSIONS ZIJIAN YAO
GALOIS EXTENSIONS ZIJIAN YAO This is a basic treatment on infinite Galois extensions, here we give necessary backgrounds for Krull topology, and describe the infinite Galois correspondence, namely, subextensions
More informationChapter 3: Baire category and open mapping theorems
MA3421 2016 17 Chapter 3: Baire category and open mapping theorems A number of the major results rely on completeness via the Baire category theorem. 3.1 The Baire category theorem 3.1.1 Definition. A
More informationCourse 311: Michaelmas Term 2005 Part III: Topics in Commutative Algebra
Course 311: Michaelmas Term 2005 Part III: Topics in Commutative Algebra D. R. Wilkins Contents 3 Topics in Commutative Algebra 2 3.1 Rings and Fields......................... 2 3.2 Ideals...............................
More informationA Crash Course in Topological Groups
A Crash Course in Topological Groups Iian B. Smythe Department of Mathematics Cornell University Olivetti Club November 8, 2011 Iian B. Smythe (Cornell) Topological Groups Nov. 8, 2011 1 / 28 Outline 1
More informationMAT 570 REAL ANALYSIS LECTURE NOTES. Contents. 1. Sets Functions Countability Axiom of choice Equivalence relations 9
MAT 570 REAL ANALYSIS LECTURE NOTES PROFESSOR: JOHN QUIGG SEMESTER: FALL 204 Contents. Sets 2 2. Functions 5 3. Countability 7 4. Axiom of choice 8 5. Equivalence relations 9 6. Real numbers 9 7. Extended
More informationBoolean Algebras, Boolean Rings and Stone s Representation Theorem
Boolean Algebras, Boolean Rings and Stone s Representation Theorem Hongtaek Jung December 27, 2017 Abstract This is a part of a supplementary note for a Logic and Set Theory course. The main goal is to
More informationFréchet algebras of finite type
Fréchet algebras of finite type MK Kopp Abstract The main objects of study in this paper are Fréchet algebras having an Arens Michael representation in which every Banach algebra is finite dimensional.
More informationVALUATION THEORY, GENERALIZED IFS ATTRACTORS AND FRACTALS
VALUATION THEORY, GENERALIZED IFS ATTRACTORS AND FRACTALS JAN DOBROWOLSKI AND FRANZ-VIKTOR KUHLMANN Abstract. Using valuation rings and valued fields as examples, we discuss in which ways the notions of
More informationFormal power series rings, inverse limits, and I-adic completions of rings
Formal power series rings, inverse limits, and I-adic completions of rings Formal semigroup rings and formal power series rings We next want to explore the notion of a (formal) power series ring in finitely
More informationCharacterized Subgroups of Topological Abelian Groups
Axioms 2015, 4, 459-491; doi:10.3390/axioms4040459 OPEN ACCESS axioms ISSN 2075-1680 www.mdpi.com/journal/axioms Article Characterized Subgroups of Topological Abelian Groups Dikran Dikranjan, Anna Giordano
More informationTHE CLOSED-POINT ZARISKI TOPOLOGY FOR IRREDUCIBLE REPRESENTATIONS. K. R. Goodearl and E. S. Letzter
THE CLOSED-POINT ZARISKI TOPOLOGY FOR IRREDUCIBLE REPRESENTATIONS K. R. Goodearl and E. S. Letzter Abstract. In previous work, the second author introduced a topology, for spaces of irreducible representations,
More informationMetric Spaces and Topology
Chapter 2 Metric Spaces and Topology From an engineering perspective, the most important way to construct a topology on a set is to define the topology in terms of a metric on the set. This approach underlies
More informationProfinite Groups. Hendrik Lenstra. 1. Introduction
Profinite Groups Hendrik Lenstra 1. Introduction We begin informally with a motivation, relating profinite groups to the p-adic numbers. Let p be a prime number, and let Z p denote the ring of p-adic integers,
More informationPart V. 17 Introduction: What are measures and why measurable sets. Lebesgue Integration Theory
Part V 7 Introduction: What are measures and why measurable sets Lebesgue Integration Theory Definition 7. (Preliminary). A measure on a set is a function :2 [ ] such that. () = 2. If { } = is a finite
More informationCHAPTER 7. Connectedness
CHAPTER 7 Connectedness 7.1. Connected topological spaces Definition 7.1. A topological space (X, T X ) is said to be connected if there is no continuous surjection f : X {0, 1} where the two point set
More informationEconomics 204 Fall 2011 Problem Set 1 Suggested Solutions
Economics 204 Fall 2011 Problem Set 1 Suggested Solutions 1. Suppose k is a positive integer. Use induction to prove the following two statements. (a) For all n N 0, the inequality (k 2 + n)! k 2n holds.
More informationCLASS NOTES FOR APRIL 14, 2000
CLASS NOTES FOR APRIL 14, 2000 Announcement: Section 1.2, Questions 3,5 have been deferred from Assignment 1 to Assignment 2. Section 1.4, Question 5 has been dropped entirely. 1. Review of Wednesday class
More informationBanach Spaces II: Elementary Banach Space Theory
BS II c Gabriel Nagy Banach Spaces II: Elementary Banach Space Theory Notes from the Functional Analysis Course (Fall 07 - Spring 08) In this section we introduce Banach spaces and examine some of their
More informationint cl int cl A = int cl A.
BAIRE CATEGORY CHRISTIAN ROSENDAL 1. THE BAIRE CATEGORY THEOREM Theorem 1 (The Baire category theorem. Let (D n n N be a countable family of dense open subsets of a Polish space X. Then n N D n is dense
More informationChapter 1. Measure Spaces. 1.1 Algebras and σ algebras of sets Notation and preliminaries
Chapter 1 Measure Spaces 1.1 Algebras and σ algebras of sets 1.1.1 Notation and preliminaries We shall denote by X a nonempty set, by P(X) the set of all parts (i.e., subsets) of X, and by the empty set.
More informationMH 7500 THEOREMS. (iii) A = A; (iv) A B = A B. Theorem 5. If {A α : α Λ} is any collection of subsets of a space X, then
MH 7500 THEOREMS Definition. A topological space is an ordered pair (X, T ), where X is a set and T is a collection of subsets of X such that (i) T and X T ; (ii) U V T whenever U, V T ; (iii) U T whenever
More informationTopology. Xiaolong Han. Department of Mathematics, California State University, Northridge, CA 91330, USA address:
Topology Xiaolong Han Department of Mathematics, California State University, Northridge, CA 91330, USA E-mail address: Xiaolong.Han@csun.edu Remark. You are entitled to a reward of 1 point toward a homework
More informationMA651 Topology. Lecture 9. Compactness 2.
MA651 Topology. Lecture 9. Compactness 2. This text is based on the following books: Topology by James Dugundgji Fundamental concepts of topology by Peter O Neil Elements of Mathematics: General Topology
More informationON THE RESIDUALITY A FINITE p-group OF HN N-EXTENSIONS
1 ON THE RESIDUALITY A FINITE p-group OF HN N-EXTENSIONS D. I. Moldavanskii arxiv:math/0701498v1 [math.gr] 18 Jan 2007 A criterion for the HNN-extension of a finite p-group to be residually a finite p-group
More informationSTRONGLY CONNECTED SPACES
Undergraduate Research Opportunity Programme in Science STRONGLY CONNECTED SPACES Submitted by Dai Bo Supervised by Dr. Wong Yan-loi Department of Mathematics National University of Singapore Academic
More informationFunctional Analysis HW #1
Functional Analysis HW #1 Sangchul Lee October 9, 2015 1 Solutions Solution of #1.1. Suppose that X
More informationCourse 212: Academic Year Section 1: Metric Spaces
Course 212: Academic Year 1991-2 Section 1: Metric Spaces D. R. Wilkins Contents 1 Metric Spaces 3 1.1 Distance Functions and Metric Spaces............. 3 1.2 Convergence and Continuity in Metric Spaces.........
More informationAlgebra SEP Solutions
Algebra SEP Solutions 17 July 2017 1. (January 2017 problem 1) For example: (a) G = Z/4Z, N = Z/2Z. More generally, G = Z/p n Z, N = Z/pZ, p any prime number, n 2. Also G = Z, N = nz for any n 2, since
More informationWeak Topologies, Reflexivity, Adjoint operators
Chapter 2 Weak Topologies, Reflexivity, Adjoint operators 2.1 Topological vector spaces and locally convex spaces Definition 2.1.1. [Topological Vector Spaces and Locally convex Spaces] Let E be a vector
More information1 Adeles over Q. 1.1 Absolute values
1 Adeles over Q 1.1 Absolute values Definition 1.1.1 (Absolute value) An absolute value on a field F is a nonnegative real valued function on F which satisfies the conditions: (i) x = 0 if and only if
More informationarxiv: v1 [math.gn] 2 Jul 2016
ON σ-countably TIGHT SPACES arxiv:1607.00517v1 [math.gn] 2 Jul 2016 ISTVÁN JUHÁSZ AND JAN VAN MILL Abstract. Extending a result of R. de la Vega, we prove that an infinite homogeneous compactum has cardinality
More information4 Countability axioms
4 COUNTABILITY AXIOMS 4 Countability axioms Definition 4.1. Let X be a topological space X is said to be first countable if for any x X, there is a countable basis for the neighborhoods of x. X is said
More informationA NEW LINDELOF SPACE WITH POINTS G δ
A NEW LINDELOF SPACE WITH POINTS G δ ALAN DOW Abstract. We prove that implies there is a zero-dimensional Hausdorff Lindelöf space of cardinality 2 ℵ1 which has points G δ. In addition, this space has
More information01. Review of metric spaces and point-set topology. 1. Euclidean spaces
(October 3, 017) 01. Review of metric spaces and point-set topology Paul Garrett garrett@math.umn.edu http://www.math.umn.edu/ garrett/ [This document is http://www.math.umn.edu/ garrett/m/real/notes 017-18/01
More information08a. Operators on Hilbert spaces. 1. Boundedness, continuity, operator norms
(February 24, 2017) 08a. Operators on Hilbert spaces Paul Garrett garrett@math.umn.edu http://www.math.umn.edu/ garrett/ [This document is http://www.math.umn.edu/ garrett/m/real/notes 2016-17/08a-ops
More informationCH AND THE MOORE-MROWKA PROBLEM
CH AND THE MOORE-MROWKA PROBLEM ALAN DOW AND TODD EISWORTH Abstract. We show that the Continuum Hypothesis is consistent with all regular spaces of hereditarily countable π-character being C-closed. This
More informationLocal Fields. Chapter Absolute Values and Discrete Valuations Definitions and Comments
Chapter 9 Local Fields The definition of global field varies in the literature, but all definitions include our primary source of examples, number fields. The other fields that are of interest in algebraic
More informationBanach Spaces V: A Closer Look at the w- and the w -Topologies
BS V c Gabriel Nagy Banach Spaces V: A Closer Look at the w- and the w -Topologies Notes from the Functional Analysis Course (Fall 07 - Spring 08) In this section we discuss two important, but highly non-trivial,
More informationFUNCTION BASES FOR TOPOLOGICAL VECTOR SPACES. Yılmaz Yılmaz
Topological Methods in Nonlinear Analysis Journal of the Juliusz Schauder Center Volume 33, 2009, 335 353 FUNCTION BASES FOR TOPOLOGICAL VECTOR SPACES Yılmaz Yılmaz Abstract. Our main interest in this
More informationMONOTONICALLY COMPACT AND MONOTONICALLY
MONOTONICALLY COMPACT AND MONOTONICALLY LINDELÖF SPACES GARY GRUENHAGE Abstract. We answer questions of Bennett, Lutzer, and Matveev by showing that any monotonically compact LOT S is metrizable, and any
More informationNotas de Aula Grupos Profinitos. Martino Garonzi. Universidade de Brasília. Primeiro semestre 2018
Notas de Aula Grupos Profinitos Martino Garonzi Universidade de Brasília Primeiro semestre 2018 1 Le risposte uccidono le domande. 2 Contents 1 Topology 4 2 Profinite spaces 6 3 Topological groups 10 4
More informationTheorem 5.3. Let E/F, E = F (u), be a simple field extension. Then u is algebraic if and only if E/F is finite. In this case, [E : F ] = deg f u.
5. Fields 5.1. Field extensions. Let F E be a subfield of the field E. We also describe this situation by saying that E is an extension field of F, and we write E/F to express this fact. If E/F is a field
More informationSolutions to Tutorial 8 (Week 9)
The University of Sydney School of Mathematics and Statistics Solutions to Tutorial 8 (Week 9) MATH3961: Metric Spaces (Advanced) Semester 1, 2018 Web Page: http://www.maths.usyd.edu.au/u/ug/sm/math3961/
More informationTopology Proceedings. COPYRIGHT c by Topology Proceedings. All rights reserved.
Topology Proceedings Web: http://topology.auburn.edu/tp/ Mail: Topology Proceedings Department of Mathematics & Statistics Auburn University, Alabama 36849, USA E-mail: topolog@auburn.edu ISSN: 0146-4124
More informationarxiv: v4 [math.gr] 2 Sep 2015
A NON-LEA SOFIC GROUP ADITI KAR AND NIKOLAY NIKOLOV arxiv:1405.1620v4 [math.gr] 2 Sep 2015 Abstract. We describe elementary examples of finitely presented sofic groups which are not residually amenable
More informationTIGHT CLOSURE IN NON EQUIDIMENSIONAL RINGS ANURAG K. SINGH
TIGHT CLOSURE IN NON EQUIDIMENSIONAL RINGS ANURAG K. SINGH 1. Introduction Throughout our discussion, all rings are commutative, Noetherian and have an identity element. The notion of the tight closure
More informationP-adic Functions - Part 1
P-adic Functions - Part 1 Nicolae Ciocan 22.11.2011 1 Locally constant functions Motivation: Another big difference between p-adic analysis and real analysis is the existence of nontrivial locally constant
More informationAN EXPLORATION OF THE METRIZABILITY OF TOPOLOGICAL SPACES
AN EXPLORATION OF THE METRIZABILITY OF TOPOLOGICAL SPACES DUSTIN HEDMARK Abstract. A study of the conditions under which a topological space is metrizable, concluding with a proof of the Nagata Smirnov
More informationLecture Notes in Advanced Calculus 1 (80315) Raz Kupferman Institute of Mathematics The Hebrew University
Lecture Notes in Advanced Calculus 1 (80315) Raz Kupferman Institute of Mathematics The Hebrew University February 7, 2007 2 Contents 1 Metric Spaces 1 1.1 Basic definitions...........................
More informationAbelian topological groups and (A/k) k. 1. Compact-discrete duality
(December 21, 2010) Abelian topological groups and (A/k) k Paul Garrett garrett@math.umn.edu http://www.math.umn.edu/ garrett/ 1. Compact-discrete duality 2. (A/k) k 3. Appendix: compact-open topology
More information10 l-adic representations
0 l-adic representations We fix a prime l. Artin representations are not enough; l-adic representations with infinite images naturally appear in geometry. Definition 0.. Let K be any field. An l-adic Galois
More informationarxiv:math/ v1 [math.gn] 28 Mar 2007
arxiv:math/0703835v1 [math.gn] 28 Mar 2007 PROJECTIVE π-character BOUNDS THE ORDER OF A π-base ISTVÁN JUHÁSZ AND ZOLTÁN SZENTMIKLÓSSY Abstract. All spaces below are Tychonov. We define the projective π-character
More informationConnectedness. Proposition 2.2. The following are equivalent for a topological space (X, T ).
Connectedness 1 Motivation Connectedness is the sort of topological property that students love. Its definition is intuitive and easy to understand, and it is a powerful tool in proofs of well-known results.
More informationSome algebraic properties of. compact topological groups
Some algebraic properties of compact topological groups 1 Compact topological groups: examples connected: S 1, circle group. SO(3, R), rotation group not connected: Every finite group, with the discrete
More informationFirst In-Class Exam Solutions Math 410, Professor David Levermore Monday, 1 October 2018
First In-Class Exam Solutions Math 40, Professor David Levermore Monday, October 208. [0] Let {b k } k N be a sequence in R and let A be a subset of R. Write the negations of the following assertions.
More informationSection II.1. Free Abelian Groups
II.1. Free Abelian Groups 1 Section II.1. Free Abelian Groups Note. This section and the next, are independent of the rest of this chapter. The primary use of the results of this chapter is in the proof
More informationNormed and Banach Spaces
(August 30, 2005) Normed and Banach Spaces Paul Garrett garrett@math.umn.edu http://www.math.umn.edu/ garrett/ We have seen that many interesting spaces of functions have natural structures of Banach spaces:
More informationON THE ALGEBRAIC DIMENSION OF BANACH SPACES OVER NON-ARCHIMEDEAN VALUED FIELDS OF ARBITRARY RANK
Proyecciones Vol. 26, N o 3, pp. 237-244, December 2007. Universidad Católica del Norte Antofagasta - Chile ON THE ALGEBRAIC DIMENSION OF BANACH SPACES OVER NON-ARCHIMEDEAN VALUED FIELDS OF ARBITRARY RANK
More information(1) A frac = b : a, b A, b 0. We can define addition and multiplication of fractions as we normally would. a b + c d
The Algebraic Method 0.1. Integral Domains. Emmy Noether and others quickly realized that the classical algebraic number theory of Dedekind could be abstracted completely. In particular, rings of integers
More informationFUNCTIONAL ANALYSIS CHRISTIAN REMLING
FUNCTIONAL ANALYSIS CHRISTIAN REMLING Contents 1. Metric and topological spaces 2 2. Banach spaces 12 3. Consequences of Baire s Theorem 30 4. Dual spaces and weak topologies 34 5. Hilbert spaces 50 6.
More informationWeek 2: Sequences and Series
QF0: Quantitative Finance August 29, 207 Week 2: Sequences and Series Facilitator: Christopher Ting AY 207/208 Mathematicians have tried in vain to this day to discover some order in the sequence of prime
More informationNOTES ON DIOPHANTINE APPROXIMATION
NOTES ON DIOPHANTINE APPROXIMATION Jan-Hendrik Evertse January 29, 200 9 p-adic Numbers Literature: N. Koblitz, p-adic Numbers, p-adic Analysis, and Zeta-Functions, 2nd edition, Graduate Texts in Mathematics
More informationFilters in Analysis and Topology
Filters in Analysis and Topology David MacIver July 1, 2004 Abstract The study of filters is a very natural way to talk about convergence in an arbitrary topological space, and carries over nicely into
More informationAxioms of separation
Axioms of separation These notes discuss the same topic as Sections 31, 32, 33, 34, 35, and also 7, 10 of Munkres book. Some notions (hereditarily normal, perfectly normal, collectionwise normal, monotonically
More information