Week 7 Exercises — solutions
Version 2024/11/20. These exercises in PDF To other course material
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Exercise 7.0. This is an unseen exercise in Applied Topology. In the diagram below, each letter of the English alphabet is drawn as a union of straight line segments and arcs.
Some letters are homeomorphic: for example, C \(\cong \) J, both are homeomorphic to a closed interval. Consider such homeomorphisms to be geometrically obvious.
Some letters are not homeomorphic: here is a topological property that can distinguish them. If \(X\) is a topological space, call \(p\in X\) a point of connectivity \(k\) if \(X\setminus \{p\}\) has exactly \(k\) connected components. The following is easy to prove: any homeomorphism \(X\xrightarrow {\sim } Y\) maps a point of connectivity \(k\) to a point of connectivity \(k.\) Hence, for each \(k,\) the number of points of connectivity \(k\) is a topological property. Example:
O \(\ncong \) P: O has no points of connectivity 2 but P has them;
T \(\ncong \) O and T \(\ncong \) P: T has a point of connectivity 3 while O, P have no such points.
CHALLENGE. Sort the letters into homeomorphism classes. You should have 9 classes.
Class 1: have a point of connectivity 4: K X
Class 2: has two points of connectivity 3: H
Class 3: one point of connectivity 3, three points of connectivity 1: E F T Y
Class 4: one point of connectivity 3, infinitely many points of connectivity 1: Q R
Class 5: one point of connectivity 2: B
Class 6: the set of points of connectivity 2 is disconnected: A
Class 7: the set of points of connectivity 2 is connected: P
Class 8: intervals — two points of connectivity 1, all other points are of connectivity 2: C G I J L M N S U V W Z
Class 9: circles — all points are of connectivity 1: D O
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Exercise 7.1. Consider the topological space \(\mathbb Q\) which is the set of all rational numbers, viewed as a subspace of the Euclidean real line \(\RR .\)
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1. Is \(\mathbb Q\) Hausdorff? Is \(\mathbb Q\) compact? Justify your answer.
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2. Show that the topology on \(\mathbb Q\) is not discrete.
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3. A topological space \(X\) is called totally disconnected if every non-empty connected subset of \(X\) is a singleton. Show that \(\mathbb Q\) is totally disconnected.
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Answer to E7.1. [These exercises without answers]
1. \(\mathbb Q\) is Hausdorff because it is a subspace of a metric (hence Hausdorff) space \(\RR .\) By Proposition 5.1, compacts in the metric space \(\mathbb R\) must be closed and bounded. Since \(\mathbb Q\) is not bounded, \(\mathbb Q\) is not compact. (Another reason for non-compactness of \(\mathbb Q\) is that \(\mathbb Q\) is not closed in \(\mathbb R.\))
2. Assume for contradiction that \(\mathbb Q\) is discrete. Then \(\{0\}\) must be an open subset of \(\mathbb Q,\) so by definition of subspace topology, \(\{0\}=\mathbb Q\cap U\) where \(U\) is open in \(\RR .\) By definition of an open set in a metric space, \(U\) must contain the open ball \((-\varepsilon ,\varepsilon )\) in \(\RR \) for some \(\varepsilon >0.\) But any interval \((-\varepsilon ,\varepsilon )\) of non-zero length contains infinitely many rational numbers, hence \(\mathbb Q\cap U\) is infinite and not \(\{0\}.\) This contradiction shows that our assumption, “\(\mathbb Q\) is discrete”, was false.
3. Let \(A\subseteq \mathbb Q\) be a non-empty connected set. The inclusion map \(\mathrm {in}\colon \mathbb Q \to \RR \) is continuous by Proposition 2.7, so \(\mathrm {in}(A)\) is an interval in \(\RR \) by Proposition 5.3. Yet \(\mathrm {in}(A)=A,\) and non-empty intervals in \(\RR \) which consist entirely of rational points are singletons (every interval of non-zero length will contain irrationals). We have proved that \(A\) is a singleton.
References for the exercise sheet
E7.1 is an enhanced version of [Armstrong, Example 3 in Section 3.5].
Version 2024/11/20. These exercises in PDF To other course material