Week 7 Exercises — solutions

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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.

    (-tikz- diagram)

    Some letters are homeomorphic: for example, C 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 pX a point of connectivity k if X{p} has exactly k connected components. The following is easy to prove: any homeomorphism XY 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:

    (-tikz- diagram)

    O P: O has no points of connectivity 2 but P has them;

    T O and T 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

  • Exercise 7.1. Consider the topological space Q which is the set of all rational numbers, viewed as a subspace of the Euclidean real line R.

    • 1. Is Q Hausdorff? Is Q compact? Justify your answer.

    • 2. Show that the topology on Q is not discrete.

    • 3. A topological space X is called totally disconnected if every non-empty connected subset of X is a singleton. Show that Q is totally disconnected.

Answer to E7.1. [These exercises without answers]

1. Q is Hausdorff because it is a subspace of a metric (hence Hausdorff) space R. By Proposition 5.1, compacts in the metric space R must be closed and bounded. Since Q is not bounded, Q is not compact. (Another reason for non-compactness of Q is that Q is not closed in R.)

2. Assume for contradiction that Q is discrete. Then {0} must be an open subset of Q, so by definition of subspace topology, {0}=QU where U is open in R. By definition of an open set in a metric space, U must contain the open ball (𝜀,𝜀) in R for some 𝜀>0. But any interval (𝜀,𝜀) of non-zero length contains infinitely many rational numbers, hence QU is infinite and not {0}. This contradiction shows that our assumption, “Q is discrete”, was false.

3. Let AQ be a non-empty connected set. The inclusion map in:QR is continuous by Proposition 2.7, so in(A) is an interval in R by Proposition 5.3. Yet in(A)=A, and non-empty intervals in R 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