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Table of Contents

  1. 8.4.3
  2. Problem 8A
  3. Problem 8B
  4. Problem 8C
  5. Problem 8D
  6. Problem 8E
  7. Problem 8F
  8. Problem 8H
  9. Problem 8I

8.4.3 #

Question 8.4.3. Give another proof using sequential definitions of continuity and compactness. (This iseven easier.)Theorem 2.3.3 (Sequential continuity)A function 𝑓:𝑀→𝑁 of metric spaces is continuous at a point 𝑝∈𝑀 if and only if the following propertyholds: if 𝑥1,𝑥2,… is a sequence in 𝑀 converging to 𝑝, then the sequence 𝑓(𝑥1),𝑓(𝑥2),… in 𝑁 converges to𝑓(𝑝).For all sequence {𝑦𝑛} in 𝑌, 𝑥𝑛=𝑓pre(𝑦𝑛) is a sequence in 𝑋.Since 𝑋 is compact, there exists a convergent subsequence {𝑥𝛾𝑘} in 𝑋.Definition 8.1.2. A metric space 𝑀 is sequentially compact if every sequence has a subsequence whichconverges.Since 𝑓 is continuous, 𝑓img(𝑥𝛾𝑘)=𝑦𝛾𝑘 is a convergent subsequence in 𝑌.Therefore, 𝑓img(𝑋) is compact.∎

Problem 8A #

Problem 8A. Show that the closed interval [0,1] and open interval (0,1) are not homeomorphic.Solution by kiwiyouSuppose [0,1] and (0,1) are homeomorphic.There exists a homeomorphism 𝑓:[0,1]→(0,1) which preserves open sets.Definition 7.2.2. A homeomorphism of topological spaces (𝑋,𝜏𝑋) and (𝑌,𝜏𝑌) is a bijection 𝑓:𝑋→𝑌 which induces a bijection from 𝜏𝑋 to 𝜏𝑌: i.e. the bijection preserves open sets.Since (0,1) is not compact, there exists an open cover {𝑈𝛼} of (0,1) without a finite subcover.Definition 8.3.2. A topological space 𝑋 is quasicompact if every open cover has a finite subcover. Itis compact if it is also Hausdorff.Since 𝑓 is a homeomorphism, {𝑓pre(𝑈𝛼)} is an open cover of [0,1], which is compact, so there exists a finitesubcover of {𝑓pre(𝑈𝛼)}, namely {𝑉𝑛}𝑁𝑛=1, which covers [0,1].Since 𝑓 is a homeomorphism, {𝑓img(𝑉𝑛)}𝑁𝑛=1 is a finite subcover of {𝑈𝛼} which covers (0,1).This is a contradiction, so [0,1] and (0,1) cannot be homeomorphic.∎

Problem 8B #

Problem 8B. Let 𝑋 be a topological space with the discrete topology. Under what conditions is 𝑋 compact?Solution by Jihyeon Kim (김지현) (simnalamburt)𝑋는 이산공간이므로 아래와 𝑋를 구성하는 집합은 아래와 같은 꼴이고𝑋={𝑥1,𝑥2,…}𝑋의 토폴로지는 아래와 같다.𝜏={∅,{𝑥1},{𝑥2},…,{𝑥1,𝑥2},{𝑥1,𝑥3},…,{𝑥1,𝑥2,𝑥3},…}𝑋 가 유한 집합이면, 𝑋는 항상 compact하다. 𝑋의 모든 open cover {𝑈𝛼}는 다음과 같은 꼴의 finite subcover를 갖기때문이다: For each 𝑥𝑖∈𝑋, we can select a 𝑉𝑖 which satisfies 𝑥𝑖∈𝑉𝑖∈{𝑈𝛼}. Then, {𝑉𝑖} is a finitesubcover of {𝑈𝛼}.𝑋 가 무한 집합이면, 𝑋는 compact하지 않다. 아래는 유효한 open cover인데, finite subcover를 갖지 않는다.{𝑈𝛼}={{𝑥}:𝑥∈𝑋}□

Problem 8C #

Problem 8C (The cofinite topology is quasicompact only). We let 𝑋 be an infinite set and equip it withthe cofinite topology: the open sets are the empty set and complements of finite sets. This makes 𝑋 intoa topological space. Show that 𝑋 is quasicompact but not Hausdorff.Solution by RanolP1.For any open cover {𝑈𝛼}, fix a non-empty open set 𝑆∈{𝑈𝛼}.Due to cofinite topology, 𝑆𝐶=𝑋∖𝑆 is finite.Let’s say |𝑆𝐶|=𝑛 and 𝑆𝐶={𝑎1,𝑎2,…,𝑎𝑛−1,𝑎𝑛}.For each 1≤𝑖≤𝑛, we pick 𝑈𝑖∈{𝑈𝛼} containing 𝑎𝑖.the set {𝑆,𝑈1,𝑈2,…,𝑈𝑛} is finite subcover for {𝑈𝛼}.We found finite subcover for all open covers.Hence 𝑋 is quasicompact. □2.For any 𝑝,𝑞∈𝑋 where 𝑝≠𝑞 and any open neighborhood of 𝑝 and 𝑞 called 𝑈 and 𝑉.If 𝑈∩𝑉=∅ then 𝑉⊆𝑋∖𝑈.However, due to cofinite topology, 𝑉 must be infinite, which cannot be a subset or equal to finite set𝑋∖𝑈.Thus 𝑈∩𝑉≠∅.Hence 𝑋 is not Hausdorff. □ReferencesDefinition 8.3.1. An open cover of a topological space 𝑋 is a collection of open sets {𝑈𝛼} (possiblyinfinite or uncountable) which cover it: every point in 𝑋 lies in at least one of the 𝑈𝛼, so that𝑋=⋃𝑈𝛼Such a cover is called an open cover.A subcover is exactly what it sounds like: it takes only some of the 𝑈𝛼, while ensuring that 𝑋 remainscovered.Definition 8.3.2. A topological space X is quasicompact if every open cover has finite subcover.Definition 7.3.1. A topological space 𝑋 is Hausdorff if for any two distinct points 𝑝 and 𝑞 in 𝑋, thereexists an open neighborhood 𝑈 of 𝑝 and an open neighborhood 𝑉 of 𝑞 such that𝑈∩𝑉=∅

Problem 8D #

Problem 8D. redactedConsider 𝐶𝑛=𝑋∖𝐾𝑛, which is open. 𝐶={𝐶𝑛|𝑛∈ℕ}.∅=𝐶0⊂𝐶1⊂𝐶2 ⋯.For any finite subset of 𝐶, the union is some 𝐶𝑛, and every 𝐶𝑛 is not 𝑋. In short, 𝐶 doesn’t have a finite subset whichcovers 𝑋.Since 𝑋 is compact and 𝐶 doesn’t have a finite subset which covers 𝑋, 𝐶 is not an open cover.Thus ∩𝑛∈ℕ𝐾𝑛≠∅. ∎

Problem 8E #

Problem 8E. redactedConsider any sequence in 𝑋×𝑌 called (𝑧𝑛).Let (𝑥𝑖𝑛) be the converging subsequence of (𝑧𝑛) projected to 𝑋. Let 𝑥 be the converging value.Let (𝑦𝑗𝑛) be the converging subsequence of (𝑧𝑖𝑛) projected to 𝑌. Let 𝑦 be the converging value.Then ∀𝜀∈ℝ+,∃𝑛1∈ℕ. ∀𝑛≥𝑛1. 𝑑(𝑥𝑗𝑛,𝑥)<𝜀2∃𝑛2∈ℕ. ∀𝑛≥𝑛2. 𝑑(𝑦𝑗𝑛,𝑦)<𝜀2∀𝑛≥max(𝑛1,𝑛2). 𝑑(𝑧𝑗𝑛,(𝑥,𝑦))<𝜀Thus 𝑧𝑗𝑛→(𝑥,𝑦). ∎

Problem 8F #

Problem 8F† (Bolzano-Weierstraß theorem for general metric spaces). Prove that a metric space 𝑀 issequentially compact if and only if it is complete and totally bounded.Solution by RanolPsequentially compact ⇒ complete and totally bounded•Given a Cauchy sequence (𝑥𝑛), because 𝑀 is sequentially compact,there exists a subsequence of (𝑥𝑛) called (𝑥𝑛𝑘) which converges to 𝑥.Claim.(𝑥𝑛)→𝑥Proof.For any 𝜀>0:∃𝑁1such that∀𝑛,𝑚≥𝑁1, 𝑑(𝑥𝑛,𝑥𝑚)<𝜀2∃𝐾such that∀𝑘≥𝐾, 𝑑(𝑥𝑛𝑘,𝑥)<𝜀2Let’s choose 𝑁=max{𝑁1,𝑛𝐾}, then for any 𝑛≥𝑁, pick a 𝑘≥𝐾 with 𝑛𝑘≥𝑛.Because of triangle inequality:𝑑(𝑥𝑛,𝑥)≤𝑑(𝑥𝑛,𝑥𝑛𝑘)+𝑑(𝑥𝑛𝑘,𝑥)<𝜀2+𝜀2=𝜀Hence (𝑥𝑛)→𝑥 □Thus, any Cauchy sequence in 𝑀 converges.Hence 𝑀 is complete □•If a metric space is sequentially compact, it is also compact (Theorem 8.3.5.). And we know that compactspace is totally bounded (Proposition 8.4.1).Hence 𝑀 is totally bounded □sequentially compact ⇐ complete and totally boundedGiven sequence (𝑥𝑛), we need to show that there exists a subsequence of (𝑥𝑛) which converges.Since 𝑀 is totally bounded, we can cover whole space with finitely many 𝜀-balls.For any 𝑘∈ℕ and any sequence (𝑦𝑛) by Infinite Pigeonhole Principle,we can pick a 1𝑘-ball called 𝐴𝑘 containing infinitely many terms in (𝑦𝑛)because 1𝑘-balls are finitely many.Let’s call the subsequence of (𝑦𝑛) contained by 𝐴𝑘 as (𝑦(𝑘)𝑛)Let’s define a subsequence (𝑥𝑛Γ𝑘) with Γ𝑘<Γ(𝑘+1) inductively for any 𝑘∈ℕ,•(𝑥𝑛Γ1)=(𝑥(1)𝑛)•(𝑥𝑛Γ(𝑘+1))=(𝑥(𝑘)𝑛Γ𝑘)Now let’s define a subsequence (𝑥𝑛𝑘) which is Cauchy.Inductively pick 𝑛1<𝑛2<⋯ with 𝑥𝑛𝑘∈(𝑥𝑛Γ𝑘).If 𝑗,𝑘≥𝑁 then 𝑥𝑛𝑗,𝑥𝑛𝑘∈𝐴𝑁 so 𝑑(𝑥𝑛𝑗,𝑥𝑛𝑘)<diameter(𝐴𝑁)≤2𝑁.Thus, for 𝜀>0, we can pick 𝑁>2𝜀 to meet 𝑑(𝑥𝑛𝑗,𝑥𝑛𝑘)<𝜀We have shown that any sequence has a Cauchy subsequence, which converges in a complete space.Hence, 𝑀 is sequentially compact □ReferencesDefinition 8.1.2. A metric space 𝑀 is sequentially compact if every sequence has a subsequencewhich converges.Theorem 8.3.5 (Sequentially compact ⟺ compact)A metric space 𝑀 is sequentially compact if and only if it is compact.Proposition 8.4.1 (Compact ⟹ totally bounded)Let 𝑀 be compact. Then 𝑀 is totally bounded.

Problem 8H #

Problem 8H. Let 𝑀=(𝑀,𝑑) be a bounded metric space. Suppose that whenever 𝑑′ is another metric on𝑀 for which (𝑀,𝑑) and (𝑀,𝑑′) are homeomorphic (i.e. have the same open sets), then 𝑑′ is also bounded.Prove that 𝑀 is compact.Solution by kiwiyouProve the contrapositive: if 𝑀 is not compact, then there exists a metric 𝑑′ on 𝑀 such that (𝑀,𝑑) and(𝑀,𝑑′) are homeomorphic, but 𝑑′ is not bounded.1. Construction of a new metricIf 𝑀 is not compact, there exists a sequence (𝑥𝑛) in 𝑀 without a convergent subsequence.Definition 8.1.2. A metric space 𝑀 is sequentially compact if every sequence has a subsequencewhich converges.Define 𝑓(𝑥)=inf𝑛(𝑑(𝑥,𝑥𝑛)+1𝑛), then 𝑓(𝑥)>0 since subsequence of (𝑥𝑛) never converges.Define 𝑑′(𝑥,𝑦)=𝑑(𝑥,𝑦)+|1𝑓(𝑥)−1𝑓(𝑦)|, then 𝑑′ is a metric.Definition 2.1.1. A metric space is a pair (𝑀,𝑑) consisting of a set of points 𝑀 and a metric 𝑑:𝑀×𝑀→ℝ≥0. The distance function must obey:•For any 𝑥,𝑦∈𝑀, we have 𝑑(𝑥,𝑦)=𝑑(𝑦,𝑥); i.e. 𝑑 is symmetric.•The function 𝑑 must be positive definite which means that 𝑑(𝑥,𝑦)≥0 with equality if and only if𝑥=𝑦.•The function 𝑑 should satisfy the triangle inequality: for all 𝑥,𝑦,𝑧∈𝑀,𝑑(𝑥,𝑧)+𝑑(𝑧,𝑦)≥𝑑(𝑥,𝑦).•𝑑′ is trivially symmetric and positive definite.•𝑑′(𝑥,𝑦) satisfies the triangle inequality.𝑑′(𝑥,𝑦)+𝑑′(𝑦,𝑧)=𝑑(𝑥,𝑦)+𝑑(𝑦,𝑧)+|1𝑓(𝑥)−1𝑓(𝑦)|+|1𝑓(𝑦)−1𝑓(𝑧)|≥𝑑(𝑥,𝑧)+|(1𝑓(𝑥)−1𝑓(𝑦))+(1𝑓(𝑦)−1𝑓(𝑧))|≥𝑑(𝑥,𝑧)+|1𝑓(𝑥)−1𝑓(𝑧)|=𝑑′(𝑥,𝑧)Therefore, 𝑑′ is a metric.2. Finding a homeomorphismNow show that (𝑀,𝑑) and (𝑀,𝑑′) are homeomorphic by identity map 𝐼.First, 𝑓 is continuous in (𝑀,𝑑).𝑑(𝑦,𝑥𝑛)≤𝑑(𝑥,𝑦)+𝑑(𝑥,𝑥𝑛)𝑑(𝑦,𝑥𝑛)+1𝑛≤𝑑(𝑥,𝑦)+𝑑(𝑥,𝑥𝑛)+1𝑛inf𝑛(𝑑(𝑦,𝑥𝑛)+1𝑛)≤𝑑(𝑥,𝑦)+inf𝑛(𝑑(𝑥,𝑥𝑛)+1𝑛)𝑓(𝑦)≤𝑑(𝑥,𝑦)+𝑓(𝑥)|𝑓(𝑦)−𝑓(𝑥)|≤𝑑(𝑥,𝑦)For all 𝜀>0,𝑑(𝑥,𝑦)<𝜀⟹|𝑓(𝑦)−𝑓(𝑥)|≤𝑑(𝑥,𝑦)<𝜀.Now that |1𝑓(𝑥)−1𝑓(𝑦)| is continuous in (𝑀,𝑑), there exists 𝛿𝑓 such that𝑑(𝑥,𝑦)<𝛿𝑓⟹|1𝑓(𝑥)−1𝑓(𝑦)|<𝜀.Define 𝛿=min(𝛿𝑓,𝜀), then𝑑(𝑥,𝑦)<𝛿⟹𝑑(𝑥,𝑦)+|1𝑓(𝑥)−1𝑓(𝑦)|<𝜀.Therefore, 𝐼 is continuous in (𝑀,𝑑).Definition 2.3.1. Let 𝑀=(𝑀,𝑑𝑀) and 𝑁=(𝑁,𝑑𝑁) be metric spaces. A function 𝑓:𝑀→𝑁 iscontinuous at a point 𝑝∈𝑀 if for every 𝜀>0 there exists a 𝛿>0 such that𝑑𝑀(𝑥,𝑝)<𝛿⟹𝑑𝑁(𝑓(𝑥),𝑓(𝑝))<𝜀.Moreover, the entire function 𝑓 is continuous if it is continuous at every point 𝑝∈𝑀.And as 𝑑(𝑥,𝑦)≤𝑑′(𝑥,𝑦), 𝐼−1 is also continuous in (𝑀,𝑑′).Since 𝐼 is a homeomorphism, so (𝑀,𝑑) and (𝑀,𝑑′) are homeomorphic.Definition 2.4.1. Let 𝑀 and 𝑁 be metric spaces. A function 𝑓:𝑀→𝑁 is a homeomorphism if itis a bijection, and both 𝑓:𝑀→𝑁 and its inverse 𝑓−1:𝑁→𝑀 are continuous. We say 𝑀 and 𝑁 arehomeomorphic.3. Unboundedness of (𝑀,𝑑′)Fix a point 𝑝, then there exists minimum integer 𝑁>1𝑓(𝑝).For all 𝑛∈ℕ, since 1𝑓(𝑥𝑛)≥𝑛,𝑑′(𝑝,𝑥𝑛+𝑁)=𝑑(𝑝,𝑥𝑛+𝑁)+|1𝑓(𝑝)−1𝑓(𝑥𝑛+𝑁)|=𝑑(𝑝,𝑥𝑛+𝑁)+1𝑓(𝑥𝑛+𝑁)−1𝑓(𝑝)≥𝑛+𝑁−𝑁≥𝑛.Therefore, 𝑑′ is not bounded.Definition 6.1.1. A metric space 𝑀 is bounded if there is a constant 𝐷 such that 𝑑(𝑝,𝑞)≤𝐷 for all𝑝,𝑞∈𝑀.Since the contrapositive is proven, 𝑀 is compact.∎

Problem 8I #

Problem 8I. In this problem a “circle” refers to the boundary of a disk with nonzero radius.(a) Is it possible to partition the plane ℝ2 into disjoint circles?(b) From the plane ℝ2 we delete two distinct points 𝑝 and 𝑞. Is it possible to partition the remaining pointsinto disjoint circles?Lemma. Suppose an open disk 𝐷⊆ℝ2 is partitioned into disjoint circles. That is, 𝐷=⨄𝐶∈𝒞𝐶 where 𝒞 is a family ofcircles in 𝐷. Then there is a sequence of circles {𝐶𝑛∈𝒞} with radius 𝑟𝑛 such that 𝑟𝑛≤(12)𝑛𝑟0.Proof. Let 𝐶0∈𝒞 be a circle whose radius is 𝑟0. By using the axiom of countable choice, it is enough to show that for all𝑛∈ℕ, there is a circle 𝐶∈𝒞 with radius 𝑟 such that 𝑟≤(12)𝑛𝑟0. We will prove this by induction on 𝑛.•Case: n = 0. 𝐶0 satisfies 𝑟0≤(12)0𝑟0.•Case: n = k + 1. Suppose there is 𝐶∈𝒞 such that 𝑟≤(12)𝑘𝑟0 by induction hypothesis. Then there is a circle 𝐶′∈𝒞containing the center of 𝐶. Let 𝑟′ be the radius of 𝐶′. Since 𝐶′ must be contained in the inside of 𝐶, 2𝑟′<𝑟. Thus𝑟′≤(12)𝑘+1𝑟0. ∎(a) No.Suppose ℝ2=⨄𝐶∈𝒞𝐶 where 𝒞 is a family of circles in ℝ2. Choose a circle 𝐶∈𝒞 and let 𝐷 be the inside of 𝐶. Then𝐷=⨄𝐶∈𝒞𝐷𝐶 where 𝒞𝐷={𝐶∈𝒞|𝐶⊆𝐷}. From the above lemma, there is a sequence of circles {𝐶𝑛∈𝒞𝐷} withradius 𝑟𝑛 and center 𝑂𝑛 such that 𝑟𝑛≤(12)𝑛𝑟0. Then 𝑂𝑛 is a cauchy sequence in ℝ2: Let 𝜀>0. We can choose 𝑁 suchthat (12)𝑁−1𝑟0<𝜀. Then for every 𝑚,𝑛≥𝑁, |𝑂𝑚−𝑂𝑛|<2𝑟𝑁≤(12)𝑁−1𝑟0<𝜀 since 𝑂𝑚,𝑂𝑛 are contained in theinside of 𝐶𝑁. Thus 𝑂𝑛→𝑂 for some 𝑂∈ℝ2. 𝑂 is contained in the inside of 𝐶𝑛 for all 𝑛. Let 𝐶∈𝒞 be the circlecontaining 𝑂 and 𝑟 be the radius of 𝐶. Then 𝐶 must be contained in the inside of 𝐶𝑛 for all 𝑛∈𝑁, implies that 𝑟<𝑟𝑛≤(12)𝑛𝑟0 for all 𝑛. But this is a contradiction since the radius of a circle must be positive. ∎(b) No.Suppose ℝ2∖{𝑝,𝑞}=⨄𝐶∈𝒞𝐶 where 𝒞 is a family of circles in ℝ2. Let 𝒫 and 𝒬 be the subset of 𝒞 such that eachcircle in 𝒫 and 𝒬 contains 𝑝 and 𝑞 in its inside, respectively. We can calssify circles in 𝒞 into one of 𝒫∩𝒬∁, 𝒫∁∩𝒬,𝒫∩𝒬, and 𝒫∁∩𝒬∁, but 𝒫∁∩𝒬∁ must be empty by the same argument of (a). Consider a strict partial order ⊏ on 𝒞such that 𝐶1⊏𝐶2 if and only if 𝐶1 is contained in the interior of 𝐶2. 𝒫∩𝒬∁, 𝒫∁∩𝒬, and 𝒫∩𝒬 are totally ordered by⊏. Let 𝑃,𝑄,𝑅 be propositions such that𝑃:𝒫∩𝒬∁is nonempty and has no maximum under⊏𝑄:𝒫∁∩𝒬is nonempty and has no maximum under⊏𝑅:𝒫∩𝒬is nonempty and has no minimum under⊏Let 𝐿 be the straight line in ℝ2 containing 𝑝 and 𝑞. 𝐿∖{𝑝,𝑞} has three connected componenet, namely 𝐿𝑝,𝐿𝑝𝑞,𝐿𝑞where 𝐿𝑝 and 𝐿𝑞 are half-line starting from 𝑝 and 𝑞, respectively, and 𝐿𝑝𝑞 is the line segment between 𝑝 and 𝑞.We claim following.1.𝑃↔¬𝑄. 𝐿𝑝𝑞=𝑋⊎𝑌 where 𝑋=𝐿𝑝𝑞∩⋃(𝒫∩𝒬∁) and 𝑌=𝐿𝑝𝑞∩⋃(𝒫∁∩𝒬).Then 𝑃↔𝑋open↔𝑌closed↔¬𝑄.2.𝑃↔¬𝑅. 𝐿𝑝=𝑋⊎𝑌 where 𝑋=𝐿𝑝∩⋃(𝒫∩𝒬∁) and 𝑌=𝐿𝑝∩⋃(𝒫∩𝒬).Then 𝑃↔𝑋open↔𝑌closed↔¬𝑅.3.𝑄↔¬𝑅. 𝐿𝑞=𝑋⊎𝑌 where 𝑋=𝐿𝑞∩⋃(𝒫∁∩𝒬) and 𝑌=𝐿𝑞∩⋃(𝒫∩𝒬).Then 𝑄↔𝑋open↔𝑌closed↔¬𝑅.Hence 𝑃↔¬𝑄↔𝑅↔¬𝑃, but this is contradiction. ∎