napkin solutions > Ch. 64

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  1. Problem 64A
  2. Problem 64B
  3. Problem 64C

Problem 64A #

Problem 64A. Show that a space 𝑋 is Hausdorff if and only if the diagonal {(𝑥,𝑥)|𝑥∈𝑋} is closed in the product space 𝑋×𝑋.Solution by RanolPLet the diagonal be 𝐷, and let 𝐷𝐶=(𝑋×𝑋)∖𝐷.𝑋 is Hausdorff ⟹ diagonal is closed in the 𝑋×𝑋For any (𝑝,𝑞)∈𝐷𝐶, we know 𝑝≠𝑞 so we can use Hausdorff.Let the open-neighborhoods of 𝑝,𝑞 be 𝑈(𝑝,𝑞),𝑉(𝑝,𝑞) where 𝑈(𝑝,𝑞)∩𝑉(𝑝,𝑞)=∅.Claim.𝐷𝐶=⋃(𝑝,𝑞)∈𝐷𝐶𝑈(𝑝,𝑞)×𝑉(𝑝,𝑞)Proof.Since 𝑈(𝑝,𝑞)∩𝑉(𝑝,𝑞)=∅, There is no 𝑧 such that (𝑧,𝑧)∈𝑈(𝑝,𝑞)×𝑉(𝑝,𝑞).In other words, it doesn’t contain any elements from 𝐷.Also, we know (𝑝,𝑞)∈𝑈(𝑝,𝑞)×𝑉(𝑝,𝑞), the union must contain every (𝑝,𝑞)∈𝐷𝐶.So the union is exactly 𝐷𝐶. □By Definition 64.3.1, 𝑈(𝑝,𝑞)×𝑉(𝑝,𝑞) is open in 𝑋×𝑋.Also (possibly infinitely many) open sets’ union is open.Hence 𝐷𝐶 is open and 𝐷 is closed. □𝑋 is Hausdorff ⟸ diagonal is closed in the 𝑋×𝑋For any (𝑥,𝑦)∈𝐷𝐶 there exists an open-neighborhood for (𝑥,𝑦) called 𝑈×𝑉.Claim.𝑈∩𝑉=∅Proof.Proof by contradiction.Suppose 𝑈∩𝑉≠∅, we can pick an element 𝑧∈𝑈∩𝑉.In other words, (𝑧,𝑧)∈𝑈×𝑉⊆𝐷𝐶.Contradiction. □For any 𝑥≠𝑦, We found disjoint open-neighborhoods for them.Hence 𝑋 is Hausdorff. □RefernecesDefinition 64.3.1. Given topological spaces 𝑋 and 𝑌, the product topology on 𝑋×𝑌 is the space whose•Points are pairs (𝑥,𝑦) with 𝑥∈𝑋,𝑦∈𝑌, and•Topology is given as follows: the basis of the topology for 𝑋×𝑌 is 𝑈×𝑉, for 𝑈⊆𝑋 open and 𝑉⊆𝑌 open.

Problem 64B #

Problem 64B. Realize the following spaces as CW complexes:(a) Möbius strip.(b) ℝ.(c) ℝ𝑛.Solution by Jihyeon Kim (김지현) (simnalamburt)(a) Let’s obtain the Möbius strip from the square [0,1]×[0,1] by identifying (1,𝑡)∼(0,1−𝑡) for all 𝑡∈[0,1].0-cells. There are two 0-cells:•𝑒0𝑎: (0,0)∼(1,1)•𝑒0𝑏: (0,1)∼(1,0)Hence 𝑋0={𝑒0𝑎,𝑒0𝑏}.1-cells. There are three 1-cells:•𝑒1: the bottom edge, from 𝑒0𝑎 to 𝑒0𝑏, (𝑡,0)•𝑓1: the top edge, from 𝑒0𝑎 to 𝑒0𝑏, (1−𝑡,1)•𝑔1: the (identified) vertical edge, from 𝑒0𝑏 to 𝑒0𝑎, (1,𝑡)∼(0,1−𝑡)So 𝑋1={𝑒1,𝑓1,𝑔1} where 𝜕𝑒1={𝑒0𝑎,𝑒0𝑏}, 𝜕𝑓1={𝑒0𝑎,𝑒0𝑏}, 𝜕𝑔1={𝑒0𝑏,𝑒0𝑎}.2-cell. Attach one 2-cell 𝜎2 whose boundary goes once around the square. Under the identification, the boundary loop is 𝑒·𝑔·𝑓·𝑔 (note 𝑔 appears twice with the same direction, creating the twist). Equivalently, attach 𝜎2 via the attaching map 𝑆1→𝑋1 that traverses the loop 𝑒𝑔𝑓𝑔.Thus 𝑋2={𝜎2} where 𝜕𝜎2=𝑒·𝑔·𝑓·𝑔.(b)𝑋0={…,𝑒0−2,𝑒0−1,𝑒00,𝑒01,𝑒02,…}𝑋1={…,𝑒1[−2,−1],𝑒1[−1,0],𝑒1[0,1],𝑒1[1,2],…} where 𝜕𝑒1[𝑛,𝑛+1]={𝑒0𝑛,𝑒0𝑛+1},(c)TBD

Problem 64C #

Problem 64C†. Show that a finite CW complex is compact.Solution by finalchildEvery 𝑘-cell is compact.Claim. Finite disjoint union of quasicompact spaces is quasicompact.Consider an open cover {𝑈𝛼} of 𝐴⊎𝐵. We denote the included indices of the finite subcovers of the projected subcovers by Β𝐴 and Β𝐵. There exists a finite subcover of {𝑈𝛼}: {𝑈𝛼|𝛼∈Β𝐴∪Β𝐵}.Claim. A quotient space of a quasicompact space is quasicompact.Consider an open cover {𝑈𝛼} of 𝑋/∼. Then, let 𝑉𝛼={𝑥|[𝑥]∈𝑈𝛼}, and {𝑉𝛼} is an open cover of 𝑋. Let Β the included indices of the finite subcover. Then {𝑈𝛼|𝛼∈Β} is the finite subcover of {𝑈𝛼}.Claim. CW complex is Hausdorff.Use mathematical induction. Only consider the case of making 𝑋𝑘 from 𝑋𝑘−1Each point of 𝑋𝑘 corresponds to either a point of 𝑋𝑘−1 (we’ll call it A class) or an inner point of an 𝑒𝑘𝛼 (we’ll call it B class).Consider two distinct points of 𝑋𝑘, 𝑎 and 𝑏.If 𝑎∈𝑋𝑘−1 and 𝑏∈inner𝑒𝑘𝛼 (or vice versa), we can take 𝑋𝑘−1 and inner𝑒𝑘𝛼 as the disjoint neighborhoods. Similarly, it is easy to take disjoint neighborhoods if one or more of 𝑎 and 𝑏 are of class B.If both 𝑎 and 𝑏 is of class A, since we assume 𝑋𝑘−1 is Hausdorff, we have disjoint open neighborhoods 𝑆𝑎 and 𝑆𝑏. Because the attaching map is continuous, the preimage of the equiv relation gives two distinct open neighborhoods in 𝑋𝑘.Thus, a finite CW complex is compact. ∎