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Normed versus topological groups: Dichotomy and duality

Normed versus topological groups: Dichotomy and duality

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<strong>Normed</strong> <strong>groups</strong> 85compact. Evidently, finite Cartesian products of shift-compact sets are shift-compact.Thus a right-shift compact set A is precompact. (If the subsequence a m t converges toa 0 t, for m in M t , then likewise a m converges to a 0 , for m in M t .)Proposition 6.6. In a normed <strong>topological</strong> group, if a subgroup S is locally right-shiftcompact, then S is closed <strong>and</strong> locally compact. Conversely, a closed, locally compact subgroupis locally right-shift compact.Proof. Suppose that a n → a 0 with a n ∈ S. If a m t → a 0 t ∈ S down a subset M, thena 0 t(a m t) −1 = a 0 a −1m ∈ S for m ∈ M. Hence also a 0 = a 0 a −1m a m ∈ S for m ∈ M. Thus Sis closed.Example. In the additive group R, the subgroup Z is closed <strong>and</strong> locally compact, soshift-compact. Of course, Z is too small to contain shifts of arbitrary null sequences. Wereturn to this matter in the remarks after Th. 7.7, where we distinguish between propershift-compactness as here (so that we are concerned only with sequences in a given set)<strong>and</strong> null-shift compactness where we are concerned with shifting subsequences of arbitrarysequences into a given set.Example. Note that A ⊆ R is density-open (open in the density topology) iff each pointof A is a density point of A. Suppose a 0 is a limit point (in the usual topology) of such aset A; then, for any ε > 0, we may find a point α ∈ A within ε/2 of a 0 <strong>and</strong> hence somet ∈ A within ε/2 of the point α such that some subsequence t + a m is included in A, withlimit t + a 0 <strong>and</strong> with |t| < ε. That is, a density-open set is strongly shift-compact.Remark. Suppose that a n = (a i n) ∈ A = ∏ A i . Pick t i <strong>and</strong> inductively infinite M i ⊆M i−1 so that a i nt i → a i 0t i along n ∈ M i with a i nt i ∈ A i for n ∈ ω. Diagonalize M i bysetting M := {m i }, where m n+1 = min{m ∈ M n+1 : m > m n }. Then the subsequence{a m : m ∈ M} satisfies, for each J finite,pr J a m t ⊆ ∏ j∈J A j for eventually all m ∈ M,<strong>and</strong> so in the product topology a m t → a 0 t through M, where (a i )(t i ) is defined to be(a i t i ).Theorem 6.7 (Shift-Compactness Theorem). In a normed <strong>topological</strong> group G, for Aprecompact, Baire <strong>and</strong> non-meagre, the set A is properly right-shift compact, i.e., for anysequence a n ∈ A, there are t ∈ G <strong>and</strong> a ∈ A such that a n t ∈ A <strong>and</strong> a n t → a down asubsequence. Likewise the set A is left-shift compact.

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