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Theorem 36.1 (Gitik). The following are equiconsistent:\n\n(i) There exists a measurable cardinal \( \kappa \) such that \( {2}^{\kappa } > {\kappa }^{ + } \).\n\n(ii) There exists a strong limit singular cardinal \( \kappa \) such that \( {2}^{\kappa } > {\kappa }^{ + } \).\n\n(iii) There exists a measurable cardinal ...
As proved in Chapter 21 (Corollary 21.13), the consistency of (ii) follows from the consistency of (i) by Prikry forcing. The necessity of (iii) for the consistency of \
No
Lemma 36.8. Assume that there exists a measurable cardinal \( \kappa \) with \( o\left( \kappa \right) = \) \( {\kappa }^{+ + } \) .\n\n(i) There is a model \( V \) that satisfies \( \mathrm{{GCH}} \) and\n\n(36.9)\n\n\[ \exists j : V \rightarrow M,\operatorname{crit}\left( j\right) = \kappa ,{M}^{\kappa } \subset M\te...
Proof. (i) We outline how to get a model that satisfies (36.9). First one uses Gitik’s forcing from [1989] to get a model of GCH that has a \( \left( {\kappa ,{\kappa }^{+ + }}\right) \) - extender \( E \) and a function \( f : \kappa \rightarrow \kappa \) such that \( j\left( f\right) \left( \kappa \right) = {\kappa }...
Yes
Lemma 36.12. \( {C}_{G} \) is a closed unbounded subset of \( \kappa \) .
Proof. Exercise 36.15.
No
Example 36.19 (Woodin). Assume that \( {\aleph }_{\omega } \) be strong limit, and let \( S \) be the following stationary set in \( P\left( {V}_{{\aleph }_{\omega + 1}}\right) \) :
\[ S = \left\{ {X \in {\left\lbrack {V}_{{\aleph }_{\omega + 1}}\right\rbrack }^{{\aleph }_{\omega }} : X \cap {\aleph }_{\omega + 1} \in {\aleph }_{\omega + 1}\text{ and }\operatorname{cf}\left( {X \cap {\aleph }_{\omega + 1}}\right) = {\aleph }_{3}}\right\} . \] Let \( G \) be a generic filter on \( {P}_{ < \delta } ...
No
Theorem 37.4 (Shelah). If \( {P}_{\alpha } \) is an RCS iteration of \( \left\{ {{\dot{Q}}_{\beta } : \beta < \alpha }\right\} \) such that every \( {\dot{Q}}_{\beta } \) is a semiproper forcing notion in \( {V}^{{P}_{\alpha } \upharpoonright \beta } \) then \( {P}_{\alpha } \) is semiproper.
Theorem 37.4 can be proved along the lines of the proof of Theorem 31.15. We shall outline the proof of a special case of Theorem 37.4 (Proposition 37.8 below) which suffices for the consistency proof of SPFA. (To be precise, Shelah proved Theorem 37.4 for a more complicated definition of RCS iteration; the current Def...
No
Proposition 37.8. If \( {P}_{\alpha } \) is an RCS iteration of semiproper forcings \( \left\{ {\dot{Q}}_{\beta }\right. \) : \( \beta < \alpha \} \) such that for every \( \beta < \alpha ,{ \Vdash }_{\beta + 1}\left| {P}_{\beta }\right| \leq {\aleph }_{1} \), then \( {P}_{\alpha } \) is semiproper.
Proof. We proceed by induction, proving (37.2). As successor stages present no problem, let \( \alpha \) be a limit ordinal. By the induction hypothesis, for every \( \gamma < \beta < \alpha ,{ \Vdash }_{\gamma }{B}_{\beta } : {B}_{\gamma } \) is semiproper; we shall prove that \( {P}_{\alpha } \) is semiproper, and (3...
No
Theorem 37.10 (Shelah). SPFA implies that every stationary set preserving notion of forcing is semiproper. Therefore SPFA implies MM.
Proof. Let \( X \) be a set of countable elementary submodels of \( {H}_{\lambda } = \left( {{H}_{\lambda }, \in , < }\right) \) . We denote \( {X}^{ \bot } \) the set\n\n(37.3)\n\n\( {X}^{ \bot } = \left\{ {M \in {\left\lbrack {H}_{\lambda }\right\rbrack }^{\omega } : M \prec {H}_{\lambda }}\right. \) and \( N \notin ...
Yes
Corollary 37.15. MM implies SCH.
Proof. For every cardinal \( \lambda \) of cofinality \( \omega \), if \( {2}^{{\aleph }_{0}} < \lambda \) then \( {\lambda }^{{\aleph }_{0}} \leq {\left( {\lambda }^{ + }\right) }^{{\aleph }_{1}} = \) \( {\lambda }^{ + } \), and SCH follows by Silver’s Theorem 8.13.
Yes
Theorem 37.16 (Foreman, Magidor and Shelah). MM implies that the nonstationary ideal on \( {\aleph }_{1} \) is \( {\aleph }_{2} \) -saturated.
Proof. Assume MM and let \( \left\{ {{A}_{i} : i \in W}\right\} \) be a maximal almost disjoint collection of stationary subsets of \( {\omega }_{1} \) . We shall find a set \( Z \subset W \) of size \( \leq {\aleph }_{1} \) such that \( \mathop{\sum }\limits_{{i \in Z}}{A}_{i} \) contains a closed unbounded set. That ...
Yes
Theorem 37.18 (Todorčević). \( \operatorname{RP}\left( {\omega }_{2}\right) \) implies that \( {2}^{{\aleph }_{0}} \leq {\aleph }_{2} \) .
Proof. One can show that \( \operatorname{RP}\left( \lambda \right) \) implies a stronger version of \( \operatorname{RP}\left( \lambda \right) \), namely that \( S \cap {\left\lbrack Y\right\rbrack }^{\omega } \) is stationary for stationary many \( Y \in {\left\lbrack \lambda \right\rbrack }^{{\aleph }_{1}} \) (Exerc...
No
Theorem 37.21. MM implies SRP.
Proof. Let \( S \subset {\left\lbrack {H}_{\kappa }\right\rbrack }^{\omega } \) be projective stationary. Let \( P \) be the forcing notion that shoots an elementary chain through \( S \) : Conditions are elementary chains \( \left\langle {{M}_{\alpha } : \alpha \leq \gamma }\right\rangle \) where \( \gamma < {\omega }...
Yes
Theorem 37.22. SRP implies that the nonstationary ideal on \( {\omega }_{1} \) is \( {\aleph }_{2} \) - saturated.
Proof. Assume SRP and let \( W \) be a maximal antichain of stationary subsets of \( {\omega }_{1} \) . We will show that \( \left| W\right| \leq {\aleph }_{1} \) . Consider the set\n\n\[ S = \left\{ {M \in {\left\lbrack {H}_{{\omega }_{2}}\right\rbrack }^{\omega } : M \prec {H}_{{\omega }_{2}}, W \in M\text{ and }\exi...
Yes
Theorem 37.23. For every regular \( \lambda \geq {\omega }_{2},\operatorname{SRP}\left( \lambda \right) \) implies \( \operatorname{RP}\left( \lambda \right) \) .
Proof. Assuming \( \operatorname{SRP}\left( \lambda \right) \) we prove a stronger version of \( \operatorname{RP}\left( \lambda \right) \) :\n\n(37.6) If \( S \) is a stationary set in \( {\left\lbrack {H}_{\lambda }\right\rbrack }^{\omega } \) then there exists an elementary chain \( \left\langle {{M}_{\alpha } : \al...
Yes
Theorem 37.26 (Shelah). MM implies \( {\mathrm{{MA}}}^{ + } \) ( \( \omega \) -closed).
Proof. Assume MM and let \( P \) be \( \omega \) -closed, \( \mathcal{D} \) a family of \( {\aleph }_{1} \) dense subsets of \( P \) and \( \dot{S} \) a \( P \) -name for a stationary set. Let \( \left\{ {{A}_{i} : i \in W}\right\} \) be a maximal antichain of those stationary sets for which \( { \Vdash }_{P}{A}_{i} \c...
Yes
Theorem 38.1 (Shelah). If there exists a Woodin cardinal then there is a generic model in which the nonstationary ideal on \( {\aleph }_{1} \) is \( {\aleph }_{2} \) -saturated.
Proof (sketch). The model is constructed by an RCS iteration (up to a Woodin cardinal), as in the proof of Theorem 37.9, iterating the forcings described in (37.5), for those maximal antichains for which the forcing (37.5) is semiproper. An argument similar to the one used in the proof of Theorem 34.8 shows that in the...
No
Theorem 38.5. The following are equiconsistent:\n\n(i) \( {I}_{\mathrm{{NS}}} \) on \( {\aleph }_{2} \) is precipitous.\n\n(ii) There exists a measurable cardinal of Mitchell order 2.
Proof. Cf. Gitik [1984]. For the lower bound, see Exercise 38.2.
No
Theorem 38.6. The following are equiconsistent:\n\n(i) There exists a cardinal \( \kappa \) such that every stationary \( S \subset \kappa \) reflects.\n\n(ii) There exists a reflecting cardinal.
Proof. Mekler and Shelah [1989].
No
If there exist infinitely many supercompact cardinals, then there is a generic model in which every stationary set \( S \subset {\aleph }_{\omega + 1} \) reflects.
Proof. Magidor [1982].
No
Theorem 38.9. The following are equiconsistent, for every \( n \geq 1 \) :\n\n(i) There exists an \( n \) -Mahlo cardinal that satisfies full reflection.\n\n(ii) There exists a \( {\Pi }_{n}^{1} \) -indescribable cardinal.
Proof. Jech and Shelah [1993]. See also Exercises 38.8 and 38.9.
No
Theorem 38.10. Let \( \kappa \) be regular uncountable and \( \lambda \geq \kappa \). (i) \( {P}_{\kappa }\left( \lambda \right) \) can be partitioned into \( \lambda \) disjoint stationary sets.
Proof. Let us consider the following set (38.1) \[ E = \left\{ {x \in {P}_{\kappa }\left( \lambda \right) : \left| {x \cap \kappa }\right| = \left| x\right| }\right\} \] It is easy to see that \( E \) is stationary and that if \( \kappa \) is a successor cardinal then \( E \) contains a closed unbounded subset (Exercis...
No
Lemma 38.11. If \( \left\{ {x \in {P}_{\kappa }\left( \lambda \right) : \left| {x \cap \kappa }\right| < \left| x\right| }\right\} \) is stationary then \( {0}^{\sharp } \) exists.
Proof. By the assumption there exists a model \( M \in {P}_{\kappa }\left( {L}_{\lambda }\right) \) such that \( \kappa \in \) \( M \prec {L}_{\lambda },{\kappa }_{M} = M \cap \kappa \in \kappa \), and \( {\kappa }_{M} < \left| M\right| \) . Let \( {L}_{\alpha } \) be the transitive collapse of \( M \) . Thus there is ...
Yes
Lemma 38.13. If \( \mathrm{{GCH}} \) holds and if \( \operatorname{cf}\lambda < \kappa \) then every stationary set in \( {P}_{\kappa }\left( \lambda \right) \) can be partitioned into \( {\lambda }^{ + } \) disjoint stationary sets.
Proof of Lemma 38.13. Assume GCH and let \( \lambda > \kappa \) be such that cf \( \lambda < \kappa \) . First we note that \( \left| {{P}_{\kappa }\left( \lambda \right) }\right| = {\lambda }^{ + } \), and that every unbounded (and therefore every stationary) subset of \( \left| {{P}_{\kappa }\left( \lambda \right) }\...
Yes
Theorem 38.14. If \( \kappa \) is a regular uncountable cardinal and \( \lambda > \kappa \) then the nonstationary ideal on \( {P}_{\kappa }\left( \lambda \right) \) is not \( {\lambda }^{ + } \) -saturated.
The result follows easily from Theorem 23.17 when \( \lambda \) is regular: Let \( \kappa < \lambda \) be regular uncountable. The proof of Theorem 23.17 shows that there are almost disjoint stationary sets \( {A}_{\xi } \subset \lambda ,\xi < {\lambda }^{ + } \), such that cf \( \alpha < \kappa \) for all \( \alpha \i...
Yes
Theorem 38.16 (Foreman-Magidor). Let \( A \) be a set of regular cardinals with \( \lambda = \sup A \). If for each \( \kappa ,{S}_{\kappa } \) is a stationary subset of \( \kappa \) such that cf \( \alpha = \omega \) for every \( \alpha \in {S}_{\kappa } \), then the \( {S}_{\kappa } \) are mutually stationary. For ev...
Proof. Foreman and Magidor [2001].
No
Corollary 38.17. If \( \lambda \) is a limit cardinal then there exist stationary stationary sets \( {S}_{\xi },\xi < {\lambda }^{\text{cf }\lambda } \), in \( {\left\lbrack \lambda \right\rbrack }^{\omega } \) such that \( {S}_{\xi } \cap {S}_{\eta } \) is nonstationary whenever \( \xi \neq \eta \) .
Proof. Let \( \mu = \operatorname{cf}\lambda \) and let \( A = \left\{ {{\kappa }_{\alpha } : \alpha < \mu }\right\} \) be a set of regular cardinals with limit \( \lambda \) . For each \( \alpha < \mu \), let \( \left\{ {{S}_{\beta }^{\alpha } : \beta < {\kappa }_{\alpha }}\right\} \) be a partition of \( {E}_{\omega ...
Yes
For every prime \( p \), the group \( {\mathbb{J}}_{p} \) of \( p \) -adic integers (see §A.4.2), considered as the ring of all endomorphisms of the Prüfer group \( \mathbb{Z}\left( {p}^{\infty }\right) \), embeds into the product \( \mathbb{Z}{\left( {p}^{\infty }\right) }^{\mathbb{Z}\left( {p}^{\infty }\right) } \) ....
The same topology on \( {\mathbb{J}}_{p} \) is induced by the product topology of \( \mathop{\prod }\limits_{{n \in \mathbb{N}}}\mathbb{Z}/{p}^{n}\mathbb{Z} \), when we consider \( {\mathbb{J}}_{p} \) as the inverse limit \( \mathop{\lim }\limits_{{ \leftarrow n \in \mathbb{N}}}\mathbb{Z}/{p}^{n}\mathbb{Z} \) (for more...
Yes
Lemma 2.1.6. Let \( G \) be a topological group. For every \( a \in G \), the left translation \( {}_{a}t : G \rightarrow \) \( G, x \mapsto {ax} \), the right translation \( {t}_{a} : G \rightarrow G, x \mapsto {xa} \), and the inner automorphism \( {\phi }_{a} : G \rightarrow G, x \mapsto {ax}{a}^{-1} \), are homeomo...
Proof. Since \( \mu : G \times G \rightarrow G,\left( {x, y}\right) \mapsto {xy} \) is continuous by Remark 2.1.2, both \( {}_{a}t = \mu \left( {a, - }\right) \) and its inverse \( {a}^{-1}t \), are continuous, for every \( a \in G \) . Hence, \( {a}^{t} \) is a homeomorphism for every \( a \in G \) . A similar proof w...
Yes
Lemma 2.1.7. Let \( G \) be a countable topological group. If \( G \) is of second category, then \( G \) is discrete. Consequently, if \( G \) is a Baire space then \( G \) is discrete.
Proof. The second assertion follows from the first, since every Baire space is of second category, by Lemma B.5.19. Assume that \( G \) is of second category. As the union \( G = \mathop{\bigcup }\limits_{{g \in G}}\{ g\} \) is countable, there exists \( g \in G \) with \( \operatorname{Int}\{ g\} \neq \varnothing \), ...
Yes
Theorem 2.1.10. Let \( G \) be a group. If \( \tau \) is a group topology on \( G \), then:\n\n(gt1) for every \( U \in {\mathcal{V}}_{\tau }\left( {e}_{G}\right) \), there exists \( V \in {\mathcal{V}}_{\tau }\left( {e}_{G}\right) \) with \( {VV} \subseteq U \) ;\n\n(gt2) for every \( U \in {\mathcal{V}}_{\tau }\left(...
Proof. Let \( \tau \) be a group topology on \( G \) . Then (gt1) and (gt2) hold by the continuity of \( \mu : G \times G \rightarrow G,\left( {x, y}\right) \mapsto {xy} \), at \( \left( {{e}_{G},{e}_{G}}\right) \) and the continuity of \( \iota : G \rightarrow G, x \mapsto {x}^{-1} \), at \( {e}_{G} \) (see Remark 2.1...
Yes
Consider the group \( \mathbb{R} \) with the Euclidean topology. Then\n\n\[ \mathcal{V}\left( 0\right) = \left\{ {U \subseteq \mathbb{R} : \exists \varepsilon \in {\mathbb{R}}_{ > 0},\left( {-\varepsilon ,\varepsilon }\right) \subseteq U}\right\} .
The base \( \mathcal{B} = \left\{ {\left( {-\varepsilon ,\varepsilon }\right) : \varepsilon \in {\mathbb{R}}_{ > 0}}\right\} \) of \( \mathcal{V}\left( 0\right) \) consisting of symmetric open neighborhoods of 0 has size \( \mathfrak{c} \) . One may choose also the countable base \( {\mathcal{B}}_{1} = \left\{ {\left( ...
Yes
Lemma 2.1.16. For a topological group \( G \) , \( \operatorname{core}\left( G\right) \) is a normal subgroup of \( G \) .
Proof. Let \( N = \operatorname{core}\left( G\right) \) . Clearly, \( {e}_{G} \in N \) . If \( x, y \in N \), then \( {xy} \in N \) by (gt1), while \( {x}^{-1} \in N \) for \( x \in N \) can be deduced from (gt2). Finally, for \( g \in G \), the inclusion \( {gN}{g}^{-1} \subseteq N \) follows from (gt3).
Yes
Lemma 2.1.20. Let \( G, H \) be topological groups and \( f : G \rightarrow H \) a homomorphism. Then the following conditions are equivalent:\n\n(a) \( f \) is continuous;\n\n(b) \( f \) is continuous at \( {e}_{G} \) ;\n\n(c) for every \( U \in {\mathcal{V}}_{H}\left( {e}_{H}\right) \) there exists \( V \in {\mathcal...
Proof. (a) \( \Leftrightarrow \) (b) immediately follows from the homogeneity of topological groups (see Exercise 2.4.1), while (c) and (d) are clearly equivalent forms of (b).
No
If \( G, H \) are Alexandrov groups, a homomorphism \( f : G \rightarrow H \) is continuous precisely when \( f\left( {\operatorname{core}\left( G\right) }\right) \subseteq \operatorname{core}\left( H\right) \).
Therefore, the category of Alexandrov groups is isomorphic to the category of pairs \( \left( {G, N}\right) \) of a group \( G \) and a normal subgroup \( N \) of \( G \), where the morphisms \( \left( {G, N}\right) \rightarrow \left( {H, L}\right) \) are group homomorphisms \( f : G \rightarrow H \) with \( f\left( N\...
No
Example 2.2.1. Let \( G \) be a group and let \( p \) be a prime. We list examples of filter bases \( \mathcal{B} \ni G \) consisting of normal subgroups of \( G \) giving rise to group topologies:
(a) the profinite topology \( {\varpi }_{G} \), with \( \mathcal{B} \) the family of all normal subgroups of \( G \) with finite index;\n\n(b) the pro-p-finite topology \( {\varpi }_{G}^{p} \), with \( \mathcal{B} \) the family of all normal subgroups of \( G \) with finite index that is a power of \( p \) ;\n\n(c) the...
Yes
Let \( p \) be a prime. The basic open neighborhoods of 0 in the topology of \( {\mathbb{J}}_{p} \) described in Example 2.1.5 are the subgroups \( {p}^{n}{\mathbb{J}}_{p} \) of \( \left( {{\mathbb{J}}_{p}, + }\right) \) for \( n \in \mathbb{N} \), that is, the topology of \( {\mathbb{J}}_{p} \) is its \( p \) -adic to...
Moreover, the \( p \) -adic topology of \( {\mathbb{J}}_{p} \) coincides with its natural topology, that is, \( {v}_{{\mathbb{J}}_{p}}^{p} = {v}_{{\mathbb{J}}_{p}} \) ; in fact, for \( m \in {\mathbb{N}}_{ + }, m{\mathbb{J}}_{p} = {p}^{s}{\mathbb{J}}_{p} \), where \( s \in \mathbb{N} \) is such that \( m = {m}_{1}{p}^{...
No
Lemma 2.2.4. Let \( G, H \) be groups and let \( f : G \rightarrow H \) be a homomorphism. Then \( f \) is continuous when both groups \( G, H \) are equipped with their profinite (respectively, pro-p-finite, pro-countable, p-adic, natural) topology.
Proof. If \( N \) is a subgroup of \( H \) of finite index, then \( {f}^{-1}\left( N\right) \) is a subgroup of finite index of \( G \) . The other cases are similar.
No
Lemma 2.2.11. Let \( G \) be an abelian group and \( H \) a subset of \( {G}^{ * } \) . The assignment \( \mathcal{P}\left( {G}^{ * }\right) \rightarrow \mathfrak{L}\left( G\right), H \mapsto {\mathcal{T}}_{H} \), is monotone increasing and \( {\mathcal{T}}_{\langle H\rangle } = {\mathcal{T}}_{H} \) .
Proof. If \( H \subseteq {H}^{\prime } \subseteq {G}^{ * } \), then \( {\mathcal{T}}_{H} \leq {\mathcal{T}}_{{H}^{\prime }} \) ; this proves the first assertion and that \( {\mathcal{T}}_{H} \leq {\mathcal{T}}_{\langle H\rangle } \) . The converse inclusion \( {\mathcal{T}}_{\langle H\rangle } \leq {\mathcal{T}}_{H} \)...
Yes
Lemma 2.2.12. Let \( G, H \) be abelian groups and let \( f : G \rightarrow H \) be a homomorphism. Then \( f : {G}^{\# } \rightarrow {H}^{\# } \) is continuous.
Proof. Let \( {\chi }_{1},\ldots ,{\chi }_{n} \in {H}^{ * } \) and \( \delta > 0 \) . Then\n\n\[ \n{f}^{-1}\left( {{U}_{H}\left( {{\chi }_{1},\ldots ,{\chi }_{n};\delta }\right) }\right) = {U}_{G}\left( {{\chi }_{1} \circ f,\ldots ,{\chi }_{n} \circ f;\delta }\right) .\n\]\n\nTo conclude apply Lemma 2.1.20.
No
Proposition 2.2.14. For every group \( G \) and every prime \( p \) , \[ {\varpi }_{G}^{p} \leq {\varpi }_{G} \leq {v}_{G} \geq {v}_{G}^{p}\;\text{ and }\;{\varpi }_{G}^{p} = \inf \{ {\varpi }_{G},{v}_{G}^{p}\} . \]
Proof. The first and the last inequality are obvious. To prove the inequality \( {\varpi }_{G} \leq {v}_{G} \) , it suffices to note that if \( N \) is a finite-index normal subgroup of \( G \), with \( \left\lbrack {G : N}\right\rbrack = m \) , then \( N \) contains the subgroup \( {M}_{m} = \left\langle {{g}^{m} : g ...
Yes
For every abelian group \( G,{\varpi }_{G} \leq \inf \left\{ {{\mathfrak{B}}_{G},{v}_{G}}\right\} \) .
In view of the inequality \( {\varpi }_{G} \leq {v}_{G} \) established in Proposition 2.2.14, it is enough to prove that \( {\varpi }_{G} \leq {\mathfrak{B}}_{G} \) .\n\nLet \( H \) be a subgroup of \( G \) of finite index; we show that \( H \) is open in \( {G}^{\# } \) . Being a finite abelian group, \( G/H \) has th...
Yes
Lemma 2.2.16. Let \( G \) be an abelian group and \( H \) a subset of \( {G}^{ * } \) . Then \( {\mathcal{T}}_{H} \leq {\varpi }_{G} \) if and only if every \( \chi \in H \) is torsion (i. e., \( H \subseteq t\left( {G}^{ * }\right) \) ).
Proof. Let \( \chi \in H \) be torsion and \( \delta > 0 \) . Then \( {U}_{G}\left( {\chi ;\delta }\right) \) contains \( \ker \chi \) . Since \( \chi \left( G\right) \cong \) \( G/\ker \chi \) is finite, \( \ker \chi \) is an open neighborhood of \( {e}_{G} \) in \( {\varpi }_{G} \), and so \( {U}_{G}\left( {\chi ;\de...
Yes
Theorem 2.2.17. For an abelian group \( G,{\mathfrak{B}}_{G} = {\varpi }_{G} \) if and only if \( G \) is bounded.
Proof. By Proposition 2.2.15, \( {\varpi }_{G} \leq {\mathfrak{B}}_{G} \) . If \( G \) is bounded, then every character of \( G \) is torsion, so Lemma 2.2.16 gives \( {\mathfrak{B}}_{G} \leq {\varpi }_{G} \) . Now assume that \( {\mathfrak{B}}_{G} = {\varpi }_{G} \) . According to Lemma 2.2.16, the group \( {G}^{ * } ...
Yes
For a nonempty set \( X \), let \( S\left( X\right) \) denote the group of all permutations of \( X \) . The stabilizer \( {S}_{x} = \{ f \in S\left( X\right) : f\left( x\right) = x\} \) of \( x \in X \) in \( S\left( X\right) \) is a subgroup of \( S\left( X\right) \) . Consider on \( S\left( X\right) \) the filter ba...
\[ {\mathcal{B}}_{X} \mathrel{\text{:=}} \left\{ {{S}_{F} : F \subseteq X\text{ finite }}\right\} ,\;\text{ where }{S}_{F} \mathrel{\text{:=}} \mathop{\bigcap }\limits_{{x \in F}}{S}_{x} = \{ f \in S\left( X\right) : \forall x \in F, f\left( x\right) = x\} . \] Since the subgroups \( {S}_{x} \) are pairwise conjugated,...
Yes
Lemma 3.1.1. Let \( G \) be a topological group. Then:\n\n(a) for a subset \( A \) of \( G \) ,\n\n\[ \bar{A} = \mathop{\bigcap }\limits_{{U \in \mathcal{V}\left( {e}_{G}\right) }}{UAU} = \mathop{\bigcap }\limits_{{U \in \mathcal{V}\left( {e}_{G}\right) }}{UA} = \mathop{\bigcap }\limits_{{U \in \mathcal{V}\left( {e}_{G...
Proof. Let \( \mathcal{V} = \mathcal{V}\left( {e}_{G}\right) \).\n\n(a) For \( x \in G \), one has \( x \notin \bar{A} \) if and only if there exists a neighborhood \( W \) of \( x \) such that \( W \cap A = \varnothing \) . Since \( G \times G \rightarrow G,\left( {a, b}\right) \mapsto {axb} \), is continuous, we can ...
Yes
Corollary 3.1.3. If \( A, B \) are nonempty subsets of a topological group \( G \) , then \( \overline{A}\overline{B} \subseteq \overline{AB} \) . If one of the sets \( A, B \) is a singleton, then \( \bar{A}\bar{B} = \overline{AB} \) .
Proof. The inclusion follows from Lemma 3.1.1(a), as \( \bar{A}\bar{B} \subseteq {UABU} \) for every \( U \in \mathcal{V}\left( {e}_{G}\right) \) . In case \( B = \{ b\} \) is a singleton, \( {AB} = {Ab} = {t}_{b}\left( A\right) \) . Since \( {t}_{b} \) is a homeomorphism by Lemma 2.1.6, \( \overline{AB} = \overline{Ab...
Yes
Proposition 3.1.7. Let \( G \) be a topological group and \( H \) a subgroup of \( G \). Then:\n\n(a) \( H \) is open in \( G \) if and only if \( H \) has nonempty interior;\n\n(b) if \( H \) is open, then \( H \) is also closed;\n\n(c) if \( \left\lbrack {G : H}\right\rbrack < \infty \), then \( H \) is closed if and...
Proof. (a) Let \( \varnothing \neq V \subseteq H \) be an open set of \( G \), and let \( {h}_{0} \in V \). Then \( {e}_{G} \in {h}_{0}^{-1}V \subseteq H = \) \( {h}_{0}^{-1}H \). Now \( U \mathrel{\text{:=}} {h}_{0}^{-1}V \subseteq H \) is open, contains \( {e}_{G} \), and \( h \in {hU} \subseteq H \) for every \( h \...
Yes
Proposition 3.1.9. Let \( G \) be an abelian group and \( H \) a subgroup of \( G \) . Then \( H \) is (dually) closed in \( \left( {G,{\mathfrak{B}}_{G}}\right) \) and \( {\left. {\mathfrak{B}}_{G}\right| }_{H} = {\mathfrak{B}}_{H} \) .
Proof. To see that \( H \) is dually closed, pick \( x \in G \smallsetminus H \) . Now it suffices to apply Corollary A.2.6 to the quotient \( G/H \) and deduce that for the nonzero element \( x + H \) of \( G/H \) there exists a character \( \xi : G/H \rightarrow \mathbb{T} \) with \( \xi \left( {x + H}\right) \neq 0 ...
Yes
Proposition 3.1.11. For a proper subgroup \( H \) of \( \mathbb{R} \), the following are equivalent:\n\n(a) \( H \) is cyclic;\n\n(b) \( H \) is discrete;\n\n(c) \( H \) is closed;\n\n(d) \( H \) is not dense in \( \mathbb{R} \).
Proof. (a) \( \Rightarrow \) (b) It is easy to see that a cyclic subgroup of \( \mathbb{R} \) is discrete.\n\n(b) \( \Rightarrow \) (c) Let \( \varepsilon > 0 \) and let the neighborhood \( U = \left( {-\varepsilon ,\varepsilon }\right) \) of 0 witness the discreteness of \( H \), that is, \( U \cap H = \{ 0\} \) . The...
Yes
According to Proposition 3.1.11, a proper subgroup of \( \mathbb{R} \) is dense if and only if it is not cyclic.
This gives easy examples of closed subgroups \( {H}_{1},{H}_{2} \) of \( \mathbb{R} \) such that \( {H}_{1} + {H}_{2} \) is not closed in \( \mathbb{R} \) . Since such \( {H}_{1},{H}_{2} \) are necessarily cyclic, we can take \( {H}_{1} = \mathbb{Z} \) and \( {H}_{2} \) any cyclic subgroup generated by an irrational nu...
Yes
Let \( {q}_{0} : \mathbb{R} \rightarrow \mathbb{T} \) be the canonical projection.\n\n(a) A proper subgroup \( H \) of \( \mathbb{T} \) is either closed (precisely when \( H \) is finite and cyclic), or dense (when \( H \) is infinite).
Indeed, \( L = {q}_{0}^{-1}\left( H\right) \) is a proper subgroup of \( \mathbb{R} \) containing \( \mathbb{Z} \) . So, if \( H \) is closed then \( L \) is closed, hence cyclic by Proposition 3.1.11 and generated by a rational since \( \mathbb{Z} \subseteq L \) ; therefore, \( H \) is a finite cyclic subgroup of \( \...
Yes
Proposition 3.1.15. Let \( G \) be a topological group. Then \( G \) is a regular topological space and the following conditions are equivalent:\n\n(a) \( G \) is \( {T}_{0} \) ;\n\n(b) \( G \) is \( {T}_{1} \) (i.e., \( \overline{\left\{ {e}_{G}\right\} } = \left\{ {e}_{G}\right\} \) );\n\n(c) \( G \) is Hausdorff;\n\...
Proof. Since \( G \) is a homogeneous topological space, to prove regularity of \( G \) it suffices to check that for every \( U \in \mathcal{V}\left( {e}_{G}\right) \) there exists \( V \in \mathcal{V}\left( {e}_{G}\right) \) such that \( \bar{V} \subseteq U \) . According to Lemma 3.1.1, it suffices to pick a \( V \i...
Yes
Proposition 3.1.17. Let \( G \) be a topological group and \( H \) a subgroup of \( G \) . If \( H \) is discrete and \( G \) is Hausdorff, then \( H \) is closed.
Proof. Since \( H \) is discrete there exists \( U \in \mathcal{V}\left( {e}_{G}\right) \) with \( U \cap H = \left\{ {e}_{G}\right\} \) . Choose \( V \in \mathcal{V}\left( {e}_{G}\right) \) with \( {V}^{-1}V \subseteq U \) . Then \( \left| {{xV} \cap H}\right| \leq 1 \) for every \( x \in G \), as \( {h}_{1} = x{v}_{1...
Yes
Proposition 3.1.20. For an infinite abelian group \( G \) and a subgroup \( H \) of \( {G}^{ * } \): (a) \( {\mathcal{T}}_{H} \) is Hausdorff if and only if the characters of \( H \) separate the points of \( G \) ; (b) \( {\mathcal{T}}_{H} \) is nondiscrete.
Proof. (a) Assume that \( H \) separates the points of \( G \) and pick a nonzero \( a \in G \) . Then there exists \( \chi \in H \) such that \( \chi \left( a\right) \neq 1 \) . Let \( \delta = \frac{1}{2}\left| {\operatorname{Arg}\left( {\chi \left( a\right) }\right) }\right| \) . Then \( U\left( {\chi ;\delta }\righ...
Yes
Proposition 3.1.21. For a group \( G \) the following conditions are equivalent:\n\n(a) the profinite topology \( {\varpi }_{G} \) of \( G \) is Hausdorff;\n\n(b) \( G \) is residually finite.\n\nIf \( G \) is abelian, then they are equivalent also to:\n\n(c) the natural topology \( {v}_{G} \) of \( G \) is Hausdorff;\...
Proof. (a) \( \Leftrightarrow \) (b) and (c) \( \Leftrightarrow \) (d) are obvious, in view of Corollary 3.1.16. Since \( {\varpi }_{G} \leq {v}_{G} \) , the first pair of conditions imply those of the second one. (d) \( \Rightarrow \) (a) follows from the fact that \( {G}^{1} \) coincides with the intersection of all ...
Yes
Theorem 3.1.22. If \( G \) is a Hausdorff group containing a dense abelian subgroup \( H \), then \( G \) is abelian.
Proof. Take \( x, y \in G \) . Then by Remark 3.1.4, there exist nets \( {\left\{ {h}_{\alpha }\right\} }_{\alpha \in A},{\left\{ {g}_{\alpha }\right\} }_{\alpha \in A} \) in \( H \) such that \( x = \mathop{\lim }\limits_{{\alpha \in A}}{h}_{\alpha }\;\mathrm{{and}}\;y = \mathop{\lim }\limits_{{\alpha \in A}}{g}_{\alp...
No
Theorem 3.1.24. If \( G \) is a Hausdorff group containing a dense nilpotent subgroup \( H \) of class \( s \), then \( G \) is nilpotent of class \( s \) .
Proof. For \( {g}_{0},{g}_{1},\ldots ,{g}_{s} \in G \), by the density of \( H \) in \( G \) and Remark 3.1.4, we can write, for every \( n \in \{ 0,1,\ldots, s\} ,{g}_{n} = \mathop{\lim }\limits_{{\alpha \in A}}{h}_{n,\alpha } \) where \( {\left\{ {h}_{n,\alpha }\right\} }_{\alpha \in A} \) is a net in \( H \) . Then,...
Yes
For a Hausdorff group \( G \), the centralizer \( {c}_{G}\left( g\right) \) of each \( g \in G \) and so the center \( Z\left( G\right) \) are closed subgroups of \( G \).
Indeed, for every \( g \in G \), by Exercise B.7.12 we conclude that \( {c}_{G}\left( g\right) \) is closed in \( G \). Then also \( Z\left( G\right) = \mathop{\bigcap }\limits_{{g \in G}}{c}_{G}\left( g\right) \) is closed in \( G \).
No
Lemma 3.2.1. Let \( G \) be a topological group, \( H \) a normal subgroup of \( G \), and let \( G/H \) be equipped with the quotient topology. Then:\n\n(a) the canonical projection \( q : G \rightarrow G/H \) is open;\n\n(b) iff: \( G/H \rightarrow {G}_{1} \) is a homomorphism to a topological group \( {G}_{1} \), th...
Proof. (a) Let \( U \neq \varnothing \) be an open set of \( G \) . Then \( {q}^{-1}\left( {q\left( U\right) }\right) = {HU} = \mathop{\bigcup }\limits_{{h \in H}}{hU} \) is open, since each \( {hU} \) is open. Therefore, \( q\left( U\right) \) is open in \( G/H \) .\n\n(b) If \( f \) is continuous, then the compositio...
Yes
Theorem 3.2.3 (Frobenius theorem). Let \( G, H \) be topological groups, \( f : G \rightarrow H \) a continuous surjective homomorphism, and \( q : G \rightarrow G/\ker f \) the canonical projection, and let \( {f}_{1} : G/\ker f \rightarrow H \) be the unique homomorphism with \( f = {f}_{1} \circ q \):\n\nThen \( {f}...
Proof. It follows immediately from the definitions of quotient topology and open map, and from Lemma 3.2.1.
No
Corollary 3.2.4. Let \( G, H \) be topological groups and \( f : G \rightarrow H \) a topological isomorphism. Then for every normal subgroup \( N \) of \( G \) the quotient \( H/f\left( N\right) \) is topologically isomorphic to \( G/N \) .
Proof. Obviously, \( f\left( N\right) \) is a normal subgroup of \( H \) and the canonical projection \( q : H \rightarrow \) \( H/f\left( N\right) \) is continuous and open by Lemma 3.2.1. Therefore, \( h = q \circ f : G \rightarrow H/f\left( N\right) \) is an open continuous surjective homomorphism with \( \ker h = N...
Yes
Proposition 3.2.5. Let \( G,{H}_{1},{H}_{2} \) be topological abelian groups and \( {f}_{i} : G \rightarrow {H}_{i}, i = 1,2 \) , open continuous surjective homomorphisms. Then there exists a continuous homomorphism \( \eta : {H}_{1} \rightarrow {H}_{2} \) such that \( {f}_{2} = \eta \circ {f}_{1} \) if and only if \( ...
Proof. The necessity is obvious. So, assume that \( \ker {f}_{1} \subseteq \ker {f}_{2} \) . By Frobenius theorem 3.2.3 applied to \( {f}_{i} \), for \( i = 1,2 \), there exists a topological isomorphism \( {j}_{i} : G/\ker {f}_{i} \rightarrow \) \( {H}_{i} \) such that \( {f}_{i} = {j}_{i} \circ {q}_{i} \), where \( {...
Yes
Lemma 3.2.6. Let \( X, Y \) be topological spaces and let \( f : X \rightarrow Y \) be an open continuous map. Then for every subspace \( P \) of \( Y \) with \( P \cap f\left( X\right) \neq \varnothing \), the restriction \( {f}^{\prime } = f{ \upharpoonright }_{{H}_{1}} : {H}_{1} \rightarrow P \) to the subspace \( {...
Proof. To see that \( {f}^{\prime } \) is open, choose a point \( x \in {H}_{1} \) and a neighborhood \( U \) of \( x \) in \( {H}_{1} \) . Then there exists a neighborhood \( W \) of \( x \) in \( X \) such that \( U = {H}_{1} \cap W \) . To see that \( {f}^{\prime }\left( U\right) \) is a neighborhood of \( {f}^{\pri...
Yes
For \( G = \mathbb{T} \) the continuous homomorphism \( {\mu }_{2} : G \rightarrow G, x \mapsto {2x} \), is surjective and open. Let now \( H = \mathbb{Z}\left( {3}^{\infty }\right) \leq \mathbb{T} \) . The restriction \( {\mu }_{2}^{\prime } : H \rightarrow {2H} = H \) of \( {\mu }_{2} \) is a continuous isomorphism.
To see that \( {\mu }_{2}^{\prime } \) is not open, it suffices to notice that the sequence \( {\left\{ {x}_{n}\right\} }_{n \in {\mathbb{N}}_{ + }} \) in \( H \) , defined by \( {x}_{n} = \mathop{\sum }\limits_{{k = 1}}^{n}1/{3}^{k} \) for every \( n \in {\mathbb{N}}_{ + } \), is not convergent in \( H \) (as it conve...
Yes
(a) If \( H \) is a subgroup of \( G \), then the homomorphism \( {q}_{1} : {HN}/N \rightarrow q\left( H\right) \), defined by \( {q}_{1}\left( {xN}\right) = q\left( x\right) \) for every \( x \in H \), is a topological isomorphism.
Proof. (a) As \( {HN} = {q}^{-1}\left( {q\left( H\right) }\right) \), we can apply Lemma 3.2.6 and conclude that the restriction \( {q}^{\prime } : {HN} \rightarrow q\left( H\right) \) of \( q \) is open. Now Frobenius theorem 3.2.3 applies to \( {q}^{\prime } \) and implies that the unique homomorphism \( {q}_{1} : {H...
Yes
Lemma 3.2.10. For a topological group \( G \) and a normal subgroup \( H \) of \( G \) :\n\n(a) \( G/H \) is discrete if and only if \( H \) is open;
Proof. Let \( q : G \rightarrow G/H \) be the canonical projection.\n\n(a) If \( G/H \) is discrete, then \( H = {q}^{-1}\left( {e}_{G/H}\right) \) is open since the singleton \( \left\{ {e}_{G/H}\right\} \) is open. If \( H \) is open, then \( \left\{ {e}_{G/H}\right\} = q\left( H\right) \) is open since the map \( q ...
Yes
Proposition 3.2.12. Let \( G, H \) be abelian groups and \( f : G \rightarrow H \) a surjective homomorphism. Then \( f \) is (continuous and) open when both \( G, H \) are equipped with their profinite (respectively, pro-p-finite, p-adic, natural, pro-countable) topology.
Proof. For the profinite and for the pro-countable topology use that fact that for a subgroup \( N \) of \( G \) the homomorphism \( {f}_{1} : G/N \rightarrow H/f\left( N\right) \) induced by \( f \) is surjective. The remaining cases are trivial.
No
Proposition 3.3.1. Let \( \left\{ {{G}_{i} : i \in I}\right\} \) be a family of topological groups. The direct product \( G = \mathop{\prod }\limits_{{i \in I}}{G}_{i} \), equipped with the product topology, is a topological group.
Proof. The filter \( \mathcal{V}\left( {e}_{G}\right) \) in the product topology of \( G \) has as a base the family of neighborhoods\n\n\[ \mathop{\bigcap }\limits_{{k = 1}}^{n}{p}_{{j}_{k}}^{-1}\left( {U}_{{j}_{k}}\right) = {U}_{{j}_{1}} \times \cdots \times {U}_{{j}_{n}} \times \mathop{\prod }\limits_{{i \in I \smal...
Yes
Proposition 3.3.6. Let \( \left\lbrack {{G}_{i},\left( {\nu }_{ij}\right), I}\right\rbrack \) be an inverse system of topological groups.\n\n(a) In the product \( H = \mathop{\prod }\limits_{{i \in I}}{G}_{i} \), consider the subgroup\n\n\[ G = \left\{ {x = {\left( {x}_{i}\right) }_{i \in I} \in H : {\nu }_{ij}\left( {...
Proof. (a) This is straightforward, since \( {p}_{i} : G \rightarrow {G}_{i} \) is continuous for every \( i \in I \) .
No
Let \( \left\{ {{G}_{n} : n \in \mathbb{N}}\right\} \) be a family of topological groups and let \( \left\{ {{\phi }_{n} : n \in \mathbb{N}}\right\} \) be a family of continuous homomorphisms \( {\phi }_{n} : {G}_{n + 1} \rightarrow {G}_{n} \) . Putting, for every pair \( m, n \in \) \( \mathbb{N} \) with \( m > n,{\va...
In these terms, if \( p \) is a prime and \( {\phi }_{n} : \mathbb{Z}\left( {p}^{n + 1}\right) \rightarrow \mathbb{Z}\left( {p}^{n}\right) \) is the canonical projection for \( n \in \mathbb{N} \), the inverse limit of the inverse system \( \left\lbrack {\mathbb{Z}\left( {p}^{n}\right) ,\left( {\phi }_{n}\right) ,\math...
Yes
Lemma 3.4.1. Let \( \left( {G,\tau }\right) \) be a topological group, \( N \mathrel{\text{:=}} \operatorname{core}\left( G\right) = \overline{\left\{ {e}_{G}\right\} } \), and \( q : G \rightarrow G/N \) the canonical projection. Then:\n\n(a) \( N \) is an indiscrete closed normal subgroup of \( G \) and \( G/N \) is ...
Proof. (a) Since \( N \subseteq U \) for every \( U \in {\mathcal{V}}_{\tau }\left( {e}_{G}\right), N \) is indiscrete. The last assertion follows from Lemma 3.2.10(b).\n\n(b) Let \( V \in {\mathcal{V}}_{\tau }\left( {e}_{G}\right) \) be open. Then \( N \subseteq V \) . Fix arbitrarily \( x \in V \) . Then there exists...
Yes
Proposition 3.4.3. Let \( G, H \) be topological groups and \( f : G \rightarrow H \) a continuous homomorphism. Then \( \mathfrak{h}f : \mathfrak{h}G \rightarrow \mathfrak{h}H \), defined by \( \mathfrak{h}f\left( {x\operatorname{core}\left( G\right) }\right) = f\left( x\right) \operatorname{core}\left( H\right) \) fo...
Proof. Since \( f\left( {e}_{G}\right) = {e}_{H} \), Proposition 3.4.2(a) implies that \( f\left( {\operatorname{core}\left( G\right) }\right) \subseteq \operatorname{core}\left( H\right) \) . This proves the correctness of the definition of \( \mathfrak{h}f \) and the commutativity of the following diagram:\n\n![71e5c...
Yes
Theorem 4.2.1. Let \( X \) be an infinite set. Then \( {\mathfrak{M}}_{S\left( X\right) } = {\mathrm{T}}_{X} \) .
This theorem follows immediately from an old result due to Gaughan:
No
Theorem 4.2.2 (Gaughan theorem). Let \( X \) be an infinite set. Every Hausdorff group topology on \( S\left( X\right) \) is finer than \( {\mathrm{T}}_{X} \) .
The proof of this theorem that we give below follows more or less the line of the proof exposed in \( \left\lbrack {{99},§{7.1}}\right\rbrack \) with several simplifications. The final stage of the proof is preceded by a number of claims (and their corollaries) and two facts about purely algebraic properties of the gro...
No
Lemma 4.2.4. For an infinite set \( X \) and \( x \in X \), the subgroup \( {S}_{x} \) of \( S\left( X\right) \) is maximal.
Proof. Assume that \( H \) is a subgroup of \( S\left( X\right) \) properly containing \( {S}_{x} \) . To show that \( H = \) \( S\left( X\right) \), take any \( f \in S\left( X\right) \) . If \( f\left( x\right) = x \), then \( f \in {S}_{x} \subseteq H \), and we are done. Assume that \( y \mathrel{\text{:=}} f\left(...
Yes
Corollary 4.2.6. If \( X \) is an infinite set and \( T \) is a Hausdorff group topology on \( S\left( X\right) \) such that \( {S}_{x} \) is \( T \) -closed in \( S\left( X\right) \) for some \( x \in X \), then \( {\mathrm{T}}_{X} \leq T \) .
Proof. Since all \( {S}_{x} \) are conjugated, the hypothesis implies that \( {S}_{x} \) is \( T \) -closed for every \( x \in X \) . By Claim 4.2.5, \( {S}_{x} \) is \( T \) -open for every \( x \in X \) . As the subgroups \( {S}_{x} \) of \( S\left( X\right) \) form a prebase of the filter of neighborhoods of \( i{d}...
Yes
Corollary 4.2.10. Let \( X \) be an infinite set and \( T \) a Hausdorff group topology on \( S\left( X\right) \) . Then for every pair \( x, y \in X \) with \( x \neq y \), the stabilizer \( {S}_{x, y} \) is not \( T \) -dense in \( S\left( X\right) \) .
Proof. By Claims 4.2.8 and 4.2.9, there exist distinct \( {x}^{\prime },{y}^{\prime } \in X \) such that \( {S}_{{x}^{\prime },{y}^{\prime }} \) is not \( T \) -dense. The assertion follows from the fact that all stabilizers of the form \( {S}_{x, y} \) with distinct \( x, y \) are conjugated.
Yes
Corollary 4.2.14. If \( G \) is a nontrivial Hausdorff group, \( X \) is an infinite set, and \( f : \left( {S\left( X\right) ,{\mathrm{T}}_{X}}\right) \rightarrow G \) is a continuous surjective homomorphism, then \( f \) is a topological isomorphism.
Proof. Since \( \left( {S\left( X\right) ,{\top }_{X}}\right) \) is topologically simple (see Exercise 8.7.7) and \( f \) is surjective and nontrivial, we deduce that \( f \) is a continuous isomorphism. Theorem 4.2.2 implies that \( f \) is open.
No
Every group with infinite center admits a nondiscrete Hausdorff group topology.
The center \( H = Z\left( G\right) \) of the group \( G \) has a nondiscrete Hausdorff group topology \( \tau \), by Proposition 3.1.20. Obviously, \( {\mathcal{V}}_{\left( H,\tau \right) }\left( {e}_{H}\right) \) is a filter base satisfying conditions (gt1), (gt2), and (gt3), so it forms a local base at \( {e}_{G} \) ...
Yes
Theorem 4.3.4. Let \( m \) and \( n \) be odd integers \( \geq {665} \) and \( A = \mathcal{A}\left( {m, n}\right) \) . The group \( G = \) \( A/{C}_{m} \) has discrete Zariski topology.
Proof. Let us see that (b), (c), and (d) jointly imply that the Zariski topology of the infinite quotient \( G = A/{C}_{m} \) is discrete (so \( G \) is a countable Markov group). Let \( d \) be a generator of \( C \) . Then \( {x}^{m} \in C \smallsetminus {C}_{m} \) for every \( x \in A \smallsetminus C \) . Indeed, i...
Yes
Under CH, there exists a group \( G \) of size \( {\omega }_{1} \) satisfying the following conditions (a) (with \( m = {10000} \) ) and (b) (with \( n = 2 \) ):\n\n(a) there exists \( m \in \mathbb{N} \) such that \( {A}^{m} = G \) for every subset \( A \) of \( G \) with \( \left| A\right| = \left| G\right| \);\n\n(b...
To see that \( G \) is a Markov group (i. e., \( {\mathfrak{M}}_{G} \) is discrete), assume that \( T \) is a Hausdorff group topology on \( G \) . There exists a \( T \) -neighborhood \( V \) of \( {e}_{G} \) with \( V \neq G \) . Choose a \( W \in {\mathcal{V}}_{T}\left( {e}_{G}\right) \) with \( \underset{m}{\underb...
Yes
Lemma 4.4.2 ([103]). Let \( H \) be a subgroup of a group \( G \) such that \( G = H{c}_{G}\left( H\right) \) . Then for every group topology \( \tau \) on \( H \), the above described topology \( {\tau }^{ * } \) is a group topology on \( G \) such that \( \left( {H,\tau }\right) \) is an open topological subgroup of ...
Proof. The first two axioms, (gt1) and (gt2), on the neighborhood base are easy to check. For (gt3), pick a basic \( {\tau }^{ * } \) -neighborhood \( U \) of \( {e}_{G} \) in \( G \) . Since \( H \) is \( {\tau }^{ * } \) -open, we can assume without loss of generality that \( U \subseteq H \), so \( U \) is a \( \tau...
Yes
Theorem 4.4.3 ([103]). Let \( H \) be a normal subgroup of the group \( G \) and let \( \tau \) be a group topology on \( H \). Then the following conditions are equivalent:\n\n(a) the extension \( {\tau }^{ * } \) is a group topology on \( G \) ;\n\n(b) \( \tau \) can be extended to a group topology of \( G \) ;\n\n(c...
Proof. (a) \( \Rightarrow \) (b) is obvious, while (b) \( \Rightarrow \) (c) follows from the fact that the conjugations are continuous in any topological group (see Lemma 2.1.6).\n\n(c) \( \Rightarrow \) (a) Assume that all automorphisms of \( H \) induced by the conjugation by elements of \( G \) are \( \tau \) -cont...
Yes
Corollary 4.4.5. For a Hausdorff group \( \left( {H,\tau }\right) \), the following conditions are equivalent:\n\n(a) every automorphism of \( H \) is \( \tau \) -continuous;\n\n(b) for every group \( G \) containing \( H \) as a normal subgroup, \( \tau \) can be extended to a group topology on G;\n\n(c) \( \tau \) ca...
Proof. (a) \( \Rightarrow \) (b) follows from Corollary 4.4.4 and (b) \( \Rightarrow \) (c) is trivial.\n\n(c) \( \Rightarrow \) (a) Extend \( \tau \) to a group topology \( {\tau }^{\prime } \) on \( G \) and note that the automorphisms of \( H \) act as restrictions of inner automorphisms of \( G \) on \( H \) . As t...
Yes
Let \( p \) be a prime number. If the group of \( p \) -adic integers \( N = {J}_{p} \) is a normal subgroup of some group \( G \), then the \( p \) -adic topology of \( N \) can be extended to a group topology on \( G \).
Indeed, it suffices to note that if \( \xi : N \rightarrow N \) is an automorphism of \( N \), then obviously \( \xi \left( {{p}^{n}N}}\right) = {p}^{n}N \) . Since the subgroups \( {p}^{n}N \) define the topology of \( N \), this proves that every automorphism of \( N \) is continuous. Now Theorem 4.4.3 applies.
Yes
Lemma 4.4.7. The only topological automorphisms \( \chi :\mathbb{T} \rightarrow \mathbb{T} \) are \( \pm i{d}_{\mathbb{T}} \) .
Proof. For \( n \in {\mathbb{N}}_{ + } \), let \( {c}_{n} = {q}_{0}\left( {1/{2}^{n}}\right) \) be the generators of \( \mathbb{Z}\left( {2}^{\infty }\right) \leq \mathbb{T} \) . Then \( {c}_{1} = {q}_{0}\left( {1/2}\right) \) is the only element of \( \mathbb{T} \) of order 2, hence \( \chi \left( {c}_{1}\right) = {c}...
Yes
Lemma 5.1.3. If \( H \) is a dense subgroup of a Hausdorff group \( G \) and \( \mathcal{B} \) is an open local base at \( {e}_{G} \) in \( H \), then \( \left\{ {{\bar{U}}^{G} : U \in \mathcal{B}}\right\} \) is a local base at \( {e}_{G} \) in \( G \) .
Proof. Since \( G \) is a regular space by Proposition 3.1.15, the closed neighborhoods of \( {e}_{G} \) in \( G \) form a local base at \( {e}_{G} \) in \( G \) . Hence, for \( V \in {\mathcal{V}}_{G}\left( {e}_{G}\right) \), one can find \( {V}_{0} \in {\mathcal{V}}_{G}\left( {e}_{G}\right) \) such that \( \overline{...
Yes
Proposition 5.1.4. Let \( H \) be a dense subset of a topological group \( G \) and let \( \mathcal{B} \) be a local base at \( {e}_{G} \) in \( G \) consisting of symmetric neighborhoods. Then \( \{ {hU} : U \in \mathcal{B}, h \in H\} \) is a base of the topology of G.
Proof. Let \( x \in G \) and let \( O \) be an open set of \( G \) containing \( x \) . Then there exists a symmetric \( U \in \mathcal{B} \) with \( {xUU} \subseteq O \) . Pick \( h \in H \cap {xU} \) . Then \( {x}^{-1}h \in U \), so \( {h}^{-1}x \in {U}^{-1} = U \) . Therefore, \( x \in {hU} = x{x}^{-1}{hU} \subseteq...
Yes
Lemma 5.1.5. Let \( G \) be a topological group. Then:\n\n(a) \( d\left( G\right) \leq w\left( G\right) \leq {2}^{d\left( G\right) } \) ;\n\n(b) in case \( G \) is Hausdorff, \( \psi \left( G\right) \leq \left| G\right| \leq {2}^{w\left( G\right) } \) .
Proof. (a) To see that \( d\left( G\right) \leq w\left( G\right) \), choose a base \( \mathcal{B} \) of the topology on \( G \) with \( \left| \mathcal{B}\right| = \) \( w\left( G\right) \), and for every \( \varnothing \neq U \in \mathcal{B} \) pick a point \( {d}_{U} \in U \) . Then the set \( D = \left\{ {{d}_{U} : ...
Yes
Lemma 5.1.7. Let \( G \) be a topological group. Then \( w\left( G\right) = \chi \left( G\right) \cdot d\left( G\right) \).
Proof. The inequality \( w\left( G\right) \geq \chi \left( G\right) \) is obvious. The inequality \( w\left( G\right) \geq d\left( G\right) \) was proved in Lemma 5.1.5(a). This gives \( w\left( G\right) \geq \chi \left( G\right) \cdot d\left( G\right) \). To prove the inequality \( w\left( G\right) \leq \chi \left( G\...
Yes
Lemma 5.1.8. Let \( H \) be a subgroup of a topological group \( G \). Then:\n\n(a) \( w\left( H\right) \leq w\left( G\right) \) and \( \chi \left( H\right) \leq \chi \left( G\right) \) ; moreover, \( \psi \left( H\right) \leq \psi \left( G\right) \) if \( G \) is \( {T}_{2} \) ;\n\n(b) if \( H \) is dense in \( G \), ...
Proof. (a) This follows from Lemma 5.1.2, as \( \left| {\{ U \cap G : U \in \mathcal{B}\} }\right| \leq \left| \mathcal{B}\right| \) for every base (respectively, local base at \( {e}_{G} \) ) \( \mathcal{B} \) of the topology on \( G \) . A similar argument applies for \( \psi \) .\n\n(b) We prove first \( \chi \left(...
Yes
Lemma 5.1.9. Let \( G, H \) be topological groups. If \( f : G \rightarrow H \) is a continuous surjective homomorphism, then \( d\left( H\right) \leq d\left( G\right) \). If \( f \) is open, then also \( w\left( H\right) \leq w\left( G\right) \) and \( \chi \left( H\right) \leq \chi \left( G\right) \).
Proof. If \( D \) is a dense subset of \( G \), then \( f\left( D\right) \) is a dense subset of \( H \) with \( \left| {f\left( D\right) }\right| \leq \left| D\right| \). This proves the first assertion. The second assertion follows from the fact that if \( \mathcal{B} \) is a base (respectively, a local base at \( {e...
Yes
Since in Lemma \( {5.1.9}\mathrm{H} \) is a continuous image of \( G \) by means of the continuous surjective homomorphism \( f : G \rightarrow H \), the canonical projection \( G \rightarrow \) \( G/\ker f \) is open, and there exists a continuous isomorphism \( G/\ker f \rightarrow H \) (which is open if and only if ...
(a) Let \( G \) be an infinite abelian group and \( {\tau }_{2} = {\delta }_{G} \), so that \( w\left( {G,{\tau }_{2}}\right) = \left| G\right| \) . Moreover, let \( {\tau }_{1} = {\mathfrak{B}}_{G} \) . Then \( w\left( {G,{\tau }_{1}}\right) = w\left( {G}^{\# }\right) = {2}^{\left| G\right| } > \left| G\right| = w\lef...
Yes
Theorem 5.1.11. If \( \left\{ {{G}_{i} : i \in I}\right\} \) is a family of topological groups and \( G = \mathop{\prod }\limits_{{i \in I}}{G}_{i} \), then \( \sup \{ d\left( {G}_{i}\right) :i \in I\} \leq d\left( G\right) \leq \left| I\right| \cdot \sup \{ d\left( {G}_{i}\right) :i \in I\} . \)
Proof. Applying Lemma 5.1.9 to the projection \( {p}_{i} : G \rightarrow {G}_{i} \), we deduce \( d\left( {G}_{i}\right) \leq d\left( G\right) \) . Hence, \( \sup \left\{ {d\left( {G}_{i}\right) : i \in I}\right\} \leq d\left( G\right) \) . \n\nFor each \( i \in I \), let \( {D}_{i} \) be a dense subset of \( {G}_{i} \...
Yes
The topological groups \( G \) with \( d\left( G\right) \leq \omega \) are precisely the separable groups.
If \( \left\{ {{G}_{i} : i \in I}\right\} \) is a family of separable groups, then \( d\left( {\mathop{\prod }\limits_{{i \in I}}{G}_{i}}\right) \leq \max \{ \omega ,\left| I\right| \} \), by Theorem 5.1.11.
No
Lemma 5.1.14. Let \( n \in {\mathbb{N}}_{ + } \), let \( {G}_{1},\ldots ,{G}_{n} \) be topological groups and \( G = {G}_{1} \times \cdots \times {G}_{n} \) . Then\n\n\[ \chi \left( G\right) = \chi \left( {G}_{1}\right) \cdots \chi \left( {G}_{n}\right) \;\text{ and }\;w\left( G\right) = w\left( {G}_{1}\right) \cdots w...
Proof. We prove the assertions for \( n = 2 \) . Then one can proceed by induction. Let \( {\mathcal{B}}_{1},{\mathcal{B}}_{2} \) be local bases at \( {e}_{{G}_{1}},{e}_{{G}_{2}} \), respectively in \( {G}_{1},{G}_{2} \) . Then\n\n\[ \mathcal{B} = \left\{ {U \times V : U \in {\mathcal{B}}_{1}, V \in {\mathcal{B}}_{2}}\...
Yes
Theorem 5.1.15. Let \( \left\{ {{G}_{i} : i \in I}\right\} \) be an infinite family of nonindiscrete topological groups and \( G = \mathop{\prod }\limits_{{i \in I}}{G}_{i} \) . Then\n\n\[ \chi \left( G\right) = \left| I\right| \cdot \sup \left\{ {\chi \left( {G}_{i}\right) : i \in I}\right\} \;\text{ and }\;w\left( G\...
Proof. For every \( i \in I \), let \( {\mathcal{B}}_{i} \) be a base of \( {\mathcal{V}}_{{G}_{i}}\left( {e}_{{G}_{i}}\right) \) with \( \left| {\mathcal{B}}_{i}\right| = \chi \left( {G}_{i}\right) \) . For any finite subset \( J \subseteq I \) and for \( {U}_{i} \in {\mathcal{B}}_{i} \) when \( i \in J \), let\n\n\[ ...
Yes
Let \( {\ell }_{2} \) denote the set of all sequences \( x = {\left\{ {x}_{n}\right\} }_{n \in \mathbb{N}} \) of real numbers such that the series \( \mathop{\sum }\limits_{{n \in \mathbb{N}}}{x}_{n}^{2} \) converges. Then \( {\ell }_{2} \) has a natural structure of vector space (induced by the Cartesian product \( {\...
This defines a norm on the abelian group \( \left( {{\ell }_{2}, + }\right) \) , that provides an invariant metric on \( {\ell }_{2} \) making it a metric space and a topological group.
Yes
Example 5.2.7. For \( n \in {\mathbb{N}}_{ + }, q \in \lbrack 1,\infty ) \) and \( x = \left( {{x}_{1},\ldots ,{x}_{n}}\right) \in {\mathbb{R}}^{n} \), let\n\n\[ \parallel x{\parallel }_{q} = {\left( \mathop{\sum }\limits_{{i = 1}}^{n}{\left| {x}_{i}\right| }^{q}\right) }^{\frac{1}{q}} \]\n\n(i) For \( q = 2 \), the no...
See Example B.3.6.
No
Let \( G \) be a finitely generated group and let \( S = {S}^{-1} \) be a finite set of generators of \( G \) . Given \( g \in G \smallsetminus \left\{ {e}_{G}\right\} \), its word norm \( {\ell }_{S}\left( g\right) \) with respect to \( S \) is the shortest length of a word in the alphabet \( S \) whose evaluation is ...
\[ {\ell }_{S}\left( g\right) = \min \left\{ {n \in {\mathbb{N}}_{ + } : g = {s}_{1}\cdots {s}_{n},{s}_{1},\ldots ,{s}_{n} \in S}\right\} . \] Moreover, let \( {\ell }_{S}\left( {e}_{G}\right) = 0 \) . It is easy to check that the function \( {\ell }_{S} \) satisfies the conditions that determine a word norm: - \( \;{\...
Yes
Theorem 5.2.14. Every Hausdorff group \( G \) is a Tichonov space.
Proof. Let \( \varnothing \neq F \) be a closed set of \( G \) with \( a \notin F \) . By the homogeneity of \( G \), we may assume that \( a = {e}_{G} \) . Then we can find a chain \( \left\{ {{U}_{n} : n \in \mathbb{N}}\right\} \) as in (5.2) of symmetric open neighborhoods of \( {e}_{G} \) with \( {U}_{n + 1}{U}_{n ...
Yes