116 N. H. Bingham <strong>and</strong> A. J. OstaszewskiRemark. Hansell’s condition, requiring the function f to be σ-discrete, is implied by fbeing analytic when X is absolutely analytic (i.e. Souslin-F(X) in any complete metricspace X into which it embeds). Frankiewicz <strong>and</strong> Kunen in [FrKu] study the consistencyrelative to ZFC of the existence of a Baire function failing to have Baire continuity.The following result provides a criterion for verifying that f is Baire.Theorem 11.9 (Souslin criterion - for Baire functions). Let X <strong>and</strong> Y be Hausdorff <strong>topological</strong>spaces with Y a K-analytic space. If f : X → Y has Souslin-F(X × Y ) graph,then f is Baire.Proof. Let G ⊆ X × Y be the graph of f which is Souslin-F(X × Y ). For F closed in Y,we havef −1 (F ) = pr X [G ∩ (X × F )],which, by the Intersection Theorem (Th. 11.3), is the projection of a Souslin-F(X × Y )set. By the Projection Theorem (Th. 11.4), f −1 (F ) is Souslin-F(X). Closed sets have theBaire property by definition, so by Nikodym’s Theorem f −1 (F ) has the Baire property.We note that in the realm of separable metric spaces, a surjective map f with analyticgraph is in fact Borel (since for U open f −1 (U) <strong>and</strong> f −1 (Y \U) are complementaryanalytic <strong>and</strong> so Borel sets, by Souslin’s Theorem (see [Jay-Rog] Th.1.4.1); for the nonseparablecase compare [Han-71, Th. 4.6(a)]).Before stating our next theorem we recall a classical result in a sharper form. We aregrateful to the referee for the statement <strong>and</strong> proof of this result in the <strong>topological</strong> <strong>groups</strong>etting, here amended to the normed group setting.Theorem 11.10 (Banach-Mehdi Continuity Theorem – [Ban-T, 1.3.4, p. 40], [Meh], [HJ,Th. 2.2.12 p. 348], or [BOst-TRII]). A Baire-continuous homomorphism f : X → Ybetween normed <strong>groups</strong>, with X Baire in the norm topology, is continuous. In particularthis is so for f Borel-measurable <strong>and</strong> Y separable.Proof. We work with the right norm topologies without loss of generality, since inversionis a homomorphism <strong>and</strong> also an isometry from the left to the right norm topology (Prop.2.5). We claim that it is enough to prove the following: for any non-empty open G inX <strong>and</strong> any ε > 0 there is a non-empty open V ⊆ G with diam(f(V )) < ε. Indeed theclaim implies that for each n ∈ N the set W n := ⋃ {V : diam(f(V )) < 1/n <strong>and</strong> V isopen <strong>and</strong> non-empty} is dense <strong>and</strong> open in X. Hence, as X is Baire, the intersection⋂n∈ W n is a non-empty set containing continuity points of f; but f is a homomorphismso is continuous everywhere.Now fix G non-empty <strong>and</strong> open <strong>and</strong> ε > 0. As f is Baire-continuous, it is continuous when
<strong>Normed</strong> <strong>groups</strong> 117restricted to X\M for some meagre set M. As M may be included in a countable unionof closed nowhere dense sets N, f restricted to some non-meagre G δ -set is continuous (e.g.to X\N). Passing to a subset, there is a non-meagre G δ -set H in X with B X ε (H) ⊆ Gsuch that diam X (H) < ε/12 <strong>and</strong> diam Y (f(H)) < ε/4.Note that HH −1 HH −1 ⊆ B X ε (e X ) (as ‖h ′ h −1 ‖ ≤ ‖h ′ ‖ + ‖h‖) <strong>and</strong> likewisef(H)f(H) −1 f(H)f(H) −1 ⊆ B Y ε/3 (e Y ). By the Squared Pettis Theorem (Th. 6.5), thereis a non-empty open set U contained in HH −1 HH −1 . Fix h ∈ H <strong>and</strong> put V := Uh. ThenV = Uh ⊆ HH −1 HH −1 h ⊆ B ε (e X )h = B ε (h) ⊆ B X ε (H) ⊆ G,<strong>and</strong> so V ⊆ G; moreover, since f is a homomorphism,f(V ) = f(Uh) = f(U)f(h) ⊆ f(H)f(H) −1 f(H)f(H) −1 f(h) ⊆ B Y ε/3 (f(h))<strong>and</strong> so diam Y f(V ) < ε, as claimed.The final assertion now follows from the Banach-Neeb Theorem (Th. 11.7).The Souslin criterion <strong>and</strong> the next theorem together have as an immediate corollarythe classical Souslin-graph Theorem; in this connection recall (see the corollary of [HJ,Th. 2.3.6 p. 355] ) that a normed group which is Baire <strong>and</strong> analytic is Polish. Ourproof, which is for normed <strong>groups</strong>, is inspired by the <strong>topological</strong> vector space proof in[Jay-Rog, §2.10] of the Souslin-graph theorem; their proof may be construed as havingtwo steps: one establishing the Souslin criterion (Th. 11.9 as above), the other the Bairehomomorphism theorem. They state without proof the <strong>topological</strong> group analogue. (See[Ost-AB] for non-separable analogues.)Theorem 11.11 (Baire Homomorphism Theorem, cf. [Jay-Rog, §2.10]). Let X <strong>and</strong> Y benormed <strong>groups</strong> with X <strong>topological</strong>ly complete. If f : X → Y is a Baire homomorphism,then f is continuous. In particular, if f is a homomorphism with a Souslin-F(X × Y )graph <strong>and</strong> Y is in addition a K-analytic space, then f is continuousProof. For f : X → Y the given homomorphism, it is enough to prove continuity at e X ,i.e. that for any ε > 0 there is δ > 0 such that B δ (e X ) ⊆ f −1 [B ε (e X )]. So let ε > 0. Wework with the right norm topology.Being K-analytic, Y is Lindelöf (cf. [Jay-Rog, Th. 2.7.1, p. 36]) <strong>and</strong> metric, so separable;so choose a countable dense set {y n } in f(X) <strong>and</strong> select a n ∈ f −1 (y n ). PutT := f −1 [B ε/4 (e Y )]. Since f is a homomorphism, f(T a n ) = f(T )f(a n ) = B ε/4 (e Y )y n .Note also that f(T −1 ) = f(T ) −1 , soT T −1 = f −1 [B ε/4 (e Y )]f −1 [B ε/4 (e Y ) −1 ] = f −1 [B ε/4 (e Y ) 2 ] ⊆ f −1 [B ε/2 (e Y )],by the triangle inequality.Nowsof(X) ⊆ ⋃ n B ε(e Y )y n ,X = f −1 (Y ) = ⋃ n T a n.
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N. H. BINGHAM and A. J. OSTASZEWSKI
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Normed groups 3ContentsContents . .
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1. IntroductionGroup-norms, which b
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Normed groups 3Topological complete
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Normed groups 5abelian group has se
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Normed groups 74 (Topological permu
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Normed groups 9The following result
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Normed groups 11Corollary 2.4. For
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Normed groups 13More generally, for
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Normed groups 15definitions, our pr
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Normed groups 17so that fg is in th
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Normed groups 19(iii) The ¯d H -to
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Normed groups 21so‖αβ‖ ≤
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Normed groups 23Remark. Note that,
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Normed groups 25shows that [z n , y
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Normed groups 27Denoting this commo
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Normed groups 29Theorem 3.4 (Equiva
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Normed groups 31argument as again p
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Normed groups 33(ii) For α ∈ H u
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Normed groups 35Definition. A group
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Normed groups 37We now give an expl
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Normed groups 39Theorem 3.19 (Abeli
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Normed groups 412. Further recall t
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Normed groups 43Theorem 3.22 (Lipsc
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Normed groups 45Proof. Z γ = G (cf
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Normed groups 47Theorem 3.30. Let G
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Normed groups 49Remark. On the matt
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Normed groups 51As for the conclusi
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Normed groups 53By (C-adm), we may
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Normed groups 55equipped with an in
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Normed groups 57Proof. To apply Th.
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Normed groups 59Definition. A point
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Normed groups 61Proposition 3.46 (M
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Normed groups 63Thus ω δ (s) ≤
- Page 69 and 70: Normed groups 65Remark. In the penu
- Page 71 and 72: Normed groups 67The result confirms
- Page 73 and 74: Normed groups 69Proof. By the Baire
- Page 75 and 76: Normed groups 715. Generic Dichotom
- Page 77 and 78: Normed groups 73Returning to the cr
- Page 79 and 80: Normed groups 75Examples. Here are
- Page 81 and 82: Normed groups 77cf. [Eng, 4.3.23].)
- Page 83 and 84: Normed groups 79Remarks. 1. See [Fo
- Page 85 and 86: Normed groups 81Theorem 6.1 (Catego
- Page 87 and 88: Normed groups 83is continuous at th
- Page 89 and 90: Normed groups 85compact. Evidently,
- Page 91 and 92: Normed groups 87j ∈ ω} which enu
- Page 93 and 94: Normed groups 89The result below ge
- Page 95 and 96: Normed groups 91left-shift, not in
- Page 97 and 98: Normed groups 93As a corollary of t
- Page 99 and 100: Normed groups 953. For X a normed g
- Page 101 and 102: Normed groups 97Proof. Note that‖
- Page 103 and 104: Normed groups 99Taking h(x) := ‖
- Page 105 and 106: Normed groups 1019. The Semigroup T
- Page 107 and 108: Normed groups 103Theorem 9.5 (Semig
- Page 109 and 110: Normed groups 105By the Category Em
- Page 111 and 112: Normed groups 107Proof. Say f is bo
- Page 113 and 114: Normed groups 109Thus G is locally
- Page 115 and 116: Normed groups 111Theorem 10.10 (Bar
- Page 117 and 118: Normed groups 113K-analyticity was
- Page 119: Normed groups 115Theorem 11.6 (Disc
- Page 123 and 124: Normed groups 119groups need not be
- Page 125 and 126: Normed groups 121Proof. In the meas
- Page 127 and 128: Normed groups 123Hence, as t i n
- Page 129 and 130: Normed groups 125The corresponding
- Page 131 and 132: Normed groups 127(t, x) ✛✻Φ T
- Page 133 and 134: Normed groups 129Fix s. Since s is
- Page 135 and 136: Normed groups 131Hence,‖x‖ −
- Page 137 and 138: Normed groups 133converging to the
- Page 139 and 140: Normed groups 135Definition. Let {
- Page 141 and 142: Normed groups 137However, whilst th
- Page 143 and 144: Normed groups 139embeddable, 14enab
- Page 145 and 146: Normed groups 141Bibliography[AL]J.
- Page 147 and 148: Normed groups 143Series 378, 2010.[
- Page 149 and 150: Normed groups 145abelian groups, Ma
- Page 151 and 152: Normed groups 147[Kak] S. Kakutani,
- Page 153 and 154: Normed groups 149fields. I. Basic p
- Page 155: Normed groups 151[So]R. M. Solovay,