Week 5 answers to exercises

Week 5 Exercises — solutions

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Exercise 5.1. Let $A$ be a subspace of a topological space $X .$ Prove: if $F \subseteq A$ and $F$ is closed in $X,$ then $F$ is closed in $A .$

Answer to E5.1. [These exercises without answers]

Assume that $F \subseteq A \subseteq X$ and that $F$ is closed as a subset of $X,$ meaning that $X ∖ F$ is open in $X .$ To show that $F$ is also closed as a subset of $A,$ we write

$A ∖ F = (X ∖ F) \cap A .$

This means that by definition of subspace topology on $A,$ the set $A ∖ F$ is open in $A :$ this is a set of the form “ $($ open in $X) \cap A$ ”. Hence $F$ is closed in $A .$

Exercise 5.2 (unions and intersections of compact sets). Let $X$ be a topological space.
- 1. Show that a union of two compact subsets of $X$ is compact.
- 2. Assuming that $X$ is Hausdorff, show that an intersection of two compact subsets of $X$ is compact. (Why do we need $X$ to be Hausdorff?)

Answer to E5.2. [These exercises without answers]

Let $K$ and $L$ be two compact subsets of $X .$

1. We use Criterion of Compactness for Subsets 4.1 to show that $M = K \cup L$ is compact. Suppose that $M$ is covered by a collection $C$ of sets open in $X .$ Then $C$ covers $K$ (because $K \subseteq M$ ). Since $K$ is compact, there exists a finite subcollection $U_{1}, \dots, U_{m} \in C$ which still covers $K .$

In the same way, there exists a finite subcollection $V_{1}, \dots, V_{n} \in C$ which covers $L .$ Then the finite subcollection $U_{1}, \dots, U_{m}, V_{1}, \dots, V_{n}$ of $C$ covers $K \cup L = M .$ By constructing a finite subcollection of $C$ which still covers $M,$ we have verified the Criterion of Compactness for $M;$ hence $M$ is compact.

2. In a Hausdorff space, a compact set is closed, Proposition 4.4, hence both $K$ and $L$ are closed in $X .$

Intersections of closed sets are closed, Proposition 2.4, so $K \cap L$ is closed in $X .$

By the previous exercise, $K \cap L$ is also closed in $K .$ A closed subset of a compact is compact, Proposition 4.3, so $K \cap L$ is compact.

Exercise 5.3 (nested sequence of closed subsets in a compact). Let $K$ be a compact topological space. Assume that $F_{1} \supseteq F_{2} \supseteq F_{3} \supseteq \dots,$ where $F_{i}$ is a non-empty closed subset of $K$ for each $i \geq 1.$ Prove that all the sets $F_{i}$ have a common point.

Answer to E5.3. [These exercises without answers]

Note that the question is about a collection of closed sets, whereas the definition of “compact” is in terms of open sets. The main idea is to pass to the complement and apply the De Morgan laws.

We are asked to prove that the intersection $⋂_{i = 1}^{\infty} F_{i}$ is not empty. Equivalently, considering the complements $U_{i} = K ∖ F_{i}$ (which are open), we need to prove that the union $⋃_{i = 1}^{\infty} U_{i}$ is not the whole of $K .$

Assume for contradiction that

$⋃_{i = 1}^{\infty} U_{i} = K .$

Then the collection $U_{1}, U_{2}, \dots$ is an open cover of $K .$ Since $K$ is compact, there is a finite subcover $U_{i_{1}}, \dots, U_{i_{n}}$ so that

$U_{i_{1}} \cup \dots \cup U_{i_{n}} = K .$

Now note that $U_{1} \subseteq U_{2} \subseteq U_{3} \subseteq \dots,$ and therefore $U_{i_{1}} \cup \dots \cup U_{i_{n}} = U_{j}$ where $j = max (i_{1}, \dots, i_{n}) .$ We have

$U_{j} = K \Rightarrow F_{j} = \emptyset .$

Yet the sets $F_{i}$ were given to be non-empty. This contradiction shows that our assumption was false.

Exercise 5.4 (a Hausdorff compact topology is “optimal”). Let $(X, T)$ be a Hausdorff compact topological space. Use the Topological Inverse Function Theorem to show that
- 1. any topology on $X,$ which is strictly weaker than $T,$ is not Hausdorff;
- 2. any topology on $X,$ which is strictly stronger than $T,$ is not compact.

Answer to E5.4. [These exercises without answers]

1. A topology $T_{w}$ on $X$ is weaker than $T$ iff the identity function ${id}_{X} : (X, T) \to (X, T_{w})$ is continuous. (We note that the function ${id}_{X}$ is always bijective.)

Assume that $(X, T)$ is compact and $(X, T_{w})$ is Hausdorff. Then by, $T$ IFT, the continuous bijection ${id}_{X} : (X, T) \to (X, T_{w})$ is a homeomorphism. That is, a set $U$ is $T$ -open iff its image ${id}_{X} (U) = U$ is $T_{w}$ -open. But this means that $T_{w} = T .$ Hence it is not possible for a Hausdorff $T_{w}$ to be strictly weaker than $T .$

Part 2. is done in a similar way and is left to the student.

References for the exercise sheet

E5.1 is a variant of [Sutherland, Exercise 10.5]. E5.2 is [Sutherland, Exercises 13.3 and 13.10]. E5.3 is [Sutherland, Exercise 13.11].

Version 2024/10/30. These exercises in PDF To other course material