Characterizations of a metrizable space X such that every An(X) is a k-space

AbstractLet X be a metrizable space, then we obtain the following results. (1) The following are equivalent:1.(a) An(X) is a k-space for each n ∈ N2.(b) A4(X) is a k-space,3.(c) the canonical mapping in : (X⊕−X⊕{0})n→An(X) is quotient for each n ∈ N4.(d) i4 is quotient,5.(e) either X is locally compact and the set X′; of all nonisolated points in X is separable, or X′ is compact. (2) The following are equivalent:1.(a) A3(X) is a k-space,2.(b) i3 is quotient,3.(c) X is locally compact or X′ is compact.(3) A2(X) is a k-space and i2 is quotient.These results contain the answers to the questions of T.H. Fay, E.T. Ordman and B.V.S. Thomas for free Abelian topological groups

Topology Proceedings

Abstract Let A(X) be the free abelian topological group on a Tychonoff space X , and S (/'l:) the sequential fan with /'l:-many spines, which is the quotient space obtained from C(/'l:) which is the disjoint union of /'l:-many con vergent sequences by identifying all the limit points to a single point. Then we proved that the tightness of A 2n (C(K)) is equal to that of S(K)n for each n E N. As a corollary, we get if /' l: is an infinite cardinal with K = W or cf(/'l:) > w, n E N and X is a metrizable space 1991 Mathematics Subject Classification: 22A05, 54A25. I(eywords and phrases: Free abelian topological group, sequential fan, tightness, maximal uniformity, pseudometric, weight of a space. 364 Kohzo Yamada such that the weight of the set X' of all non-isolated points in X is K, then t(S(K)n) = t(A 2n (C(K))) ~ t(A 2n (X)) ~ t(A(X)) ~ w(X') = K. This result partially answers the question raised by Arhangel'skir, Okunev and Pestov in o Introduction This paper discusses the relations between the tightness of free abelian topological groups A(X) on metrizable spaces X, the tightness of the subspaces An(X) of A(X) formed by reduced words whose lengths are less than or equal to n, n == 1, 2, ... , and the tightness of finite products of sequential fans S(K). The tightness of a space X is denoted by t(X) and the weight of X is denoted by w(X). The set of all natural num bers is denoted by N. For an infinite cardinal K, let C (K) be the topological sum of K-many convergent sequences with their limits points. Then we prove that t(A 2n (C(K))) == t(S(K)n) for each n E N. In Let K be an infinite cardinal such that its cofinality cf( K) is larger than wand X a metrizable space such that w(X') == K. Ift(S(K)n) == K for some n E N, then t(A 2n (X)) == t(A(X)) == w(X') == K. On the other hand, in the same paper Arhangel'skil, Okunev and Pestov proved that if t(A(X)) == w, the converse inequal ity is true, i.e. t(A(X)) == w(X'). Since Gruenhage and (2) X and An(X)J n E N J are closed subspaces of A(X). On the other hand, using the universal uniformity U x of X, another representation of a neighborhood base of 0 in A(X) was given in Furthermore, as a corollary of the above Lemma, we have the following, which is often used in our main results. For a space X and each n E N, we define a mapping jn from x 2 n to A 2n (X) as follows; (Xl,X2, ... ,X n ) and (Yl,Y2, ... ,Yn) in xn. Tightness of A 2 (X) For a space X, since A1(X) = X EB -X EB {O}, it easy to see that t(AI(X)) = t(X). For the tightness of A 2 (X), we can show the following. Since t(X EB -X) :::; K and by the property (5) in Lemma 1.1, we can take a subset C of E such that 9 E C and ICI :::; K. The proof is in two cases. In this case, it is easy to see that Then we can take an open set U in X 2 and an open set V in Al (X). By the property (5) in Lemma 1.1, we have that t(U) :::; K, and hence we can take a subset C of F such that ~ E C and lei:::; K. is homeomorphic to X and X is a subset of A 2 (X). It follows that t(~x) :::; K. Then, we may assume that z E F, where We define a subset A of D.x, as follows; A == {y E ~x: there is C y C F such that y E C y and ICyl :::; K}. Since z E ~x, F n ~x # 0, and hence 0 E }l (F). . Then t(A 2 (X)) = t(X 2 ). The following example shows that the hypothesis of a space X in Theorem 2.1 cannot be omitted. It was proved by Ohta and is presented here with his kind permission. Since C is closed unbounded in WI, C can be decomposed into two disjoint stationary sets C l = {a" : , E WI} and C2 = {(3" : , E WI}, where a" < a,,' and (3" < (3", if , < ,'. Example 2.3 There exists a space Since {j( a'Y(n)) }nEw and {j ((3"Y(n)) }nEw have the same limit, there is mEw such that d(p, q) < 1/3, where p == (v,j(a'Y(m))) and q == (v,!((3'Y(m) D Let X be a space and fix an n E N. When we take a subset H of An(X) and a word 9 E An(X) such that 9 E H to investigate the tightness of An(X), by Lemma 1.1, it suffices to consider the following three cases: for some k ::; n. (2) 9 == 0 and 9 E H n (A 2k (X) \ A 2k -1 (X)) for some k with 2k ::; n. (3) 9 E Ak(X) \ Ak-1 (X) and 9 E H n (A2m+ k (X) \ A 2m +k-1 (X)) for some k and m with 2m + k ~ n. Tightness of Free Abelian Topological Groups 371 In the case (1), We can apply the property where E == {eA == Proof: We prove the Theorem by induction with respect to n. It is clear that t(S(K)) = w and, by Theorem 2.1, we have that Claim 1. t(A 2n (C(K))) ~ t(S(K)n). Let t(S(K)n) = T, (w ~ T ~ K), and take a subset H of A 2n (C(K)) and a word 9 E A 2n (C(K)) such that 9 E H. By our inductive assumption, we may assume that by the properties (3) and (5) in Lemma 1.1, we can take a countable subset H o of H such that 9 E H o . Thus, it suffices to show in the following two cases. . Thus, by Lemma 3.1, we can take a subset H o of H such that 9 E H o and Similarly, if k is even, we can take a subset H o of H such that 9 E H o and IHol ::; t(A 2n -k (C(K))). In any case, by our inductive assumption, we have that IHol ::; Case 2: 9 = O. 376 Kohzo Yamada In this case, we may assume that (1) t(S(K)n):::; t(A 2n (C(K))). Let t (A 2n (C(K) From Claim 1 and 2, we can prove that t(A 2n (C(K))) t(S(K)n). 0 Arhangel'skil, Okunev and Pestov t(S(K)n) == t(A 2n (C(K))) :::; t(A(C(K)) :::; W(C(K)') == K. Let X be a metrizable space with w(X') == K, where Cf(K) > w. Then C(K) can be embedded in X as a closed subset. By this fact, the property (6) of Lemma 1.1 and Corollary 3.3, we obtain the following. where Cf(K) > w. Assume that t(S(K)n) = K for some n E N. Then t(A 2n (X)) Corollary 3.6 Let X be a metrizable space with w(X ' ) = WI. Then t(A 4 (X)) = t(A(X)) = w(X ' ) = WI. Proof: Since Gruenhage and is a k-space and i 2 is quotient. On the other hand, apply the results in the previous section. Then, with respect to the resul

Products of straight spaces with compact spaces

[EN] A metric space X is called straight if any continuous real-valued function which is uniformly continuous on each set of a finite cover of X by closed sets, is itself uniformly continuous. Let C be the convergent sequence {1/n : n ϵ N} with its limit 0 in the real line with the usual metric. In this paper, we show that for a straight space X, X × C is straight if and only if X × K is straight for any compact metric space K. Furthermore, we show that for a straight space X, if X × C is straight, then X is precompact. Note that the notion of straightness depends on the metric on X. Indeed, since the real line R with the usual metric is not precompact, R×C is not straight. On the other hand, we show that the product space of an open interval and C is straight.Nishijima, K.; Yamada, K. (2007). Products of straight spaces with compact spaces. Applied General Topology. 8(2):151-159. doi:10.4995/agt.2007.1877.SWORD1511598

Free topological groups on metrizable spaces and inductive limits

AbstractWe prove that for a metrizable space X the following are equivalent: (i) the free Abelian topological group A(X) is the inductive limit of the sequence {An(X):n∈N}, where An(X) is formed by all words of reduced length ≤n; (ii) X is locally compact and the set of all non-isolated points of X is separable. In the non-Abelian case, for a metrizable X the following are equivalent: (i) the free topological group F(X) is the inductive limit of the sequence {Fn(X):n∈N}; (ii) X is either locally compact separable or discrete

Myocardial velocity gradient as a noninvasively determined index of left ventricular diastolic dysfunction in patients with hypertrophic cardiomyopathy

AbstractObjectivesWe investigated the utility of the peak negative myocardial velocity gradient (MVG) derived from tissue Doppler imaging (TDI) for evaluation of diastolic dysfunction in patients with hypertrophic cardiomyopathy (HCM).BackgroundHypertrophic cardiomyopathy is characterized by impaired diastolic function with abnormal stiffness and prolonged relaxation. However, it remains difficult to evaluate these defects noninvasively.MethodsBoth TDI and conventional echocardiography were performed in 36 patients with HCM and in 47 control subjects. Left ventricular (LV) pressure was measured simultaneously in all HCM patients and in 26 controls.ResultsThe peak negative MVG occurred soon after the isovolumic relaxation period during the initial phase of rapid filling (auxotonic relaxation). It was significantly smaller in HCM patients than in control subjects (2.32 ± 0.52/s vs. 4.82 ± 1.15/s, p < 0.0001); the cutoff value for differentiation between all HCM patients and 47 normal individuals was determined as 3.2/s. Both the left ventricular end-diastolic pressure (LVEDP) (19.6 ± 6.1 mm Hg vs. 6.5 ± 1.7 mm Hg, p < 0.0001) and the time constant of LV pressure decay during isovolumic diastole (tau) (44.0 ± 6.7 ms vs. 32.1 ± 5.5 ms, p < 0.0001) were increased in HCM patients compared with controls. The peak negative MVG was negatively correlated with both LVEDP (r= −0.75, p < 0.0001) and tau (r= −0.58, p < 0.0001).ConclusionsA reduced peak negative MVG reflects both prolonged relaxation and elevated LVEDP. The peak negative MVG might thus provide a noninvasive index of diastolic function, yielding unique information about auxotonic relaxation in patients with HCM

Pelvic actinomycosis presenting with a large abscess and bowel stenosis with marked response to conservative treatment: a case report

Pelvic actinomycosis is a rare disease that can result in abscess formation, bowel obstruction, and other serious complications. Moreover, the correct diagnosis can seldom be established before radical surgery because the disease often mimics pelvic neoplasms. It has been recently recognized that pelvic actinomycosis is associated with long-term use of an intrauterine contraceptive device