Question

(a) Show that a closed subset of a compact set is compact. (b) Suppose that \(\mathcal{T}\) is any collection of closed subsets of a metric space \((A, \rho)\). and some \(\widehat{T}\) in \(\mathcal{T}\) is compact. Show that \(\cap\\{T \mid T \in \mathcal{T}\\}\) is compact. (c) Show that if \(T\) is a collection of compact subsets of a metric space \((A, \rho)\), then \(\cap\\{T \mid T \in \mathcal{T}\\}\) is compact.

Short answer

#Question# Show that a closed subset of a compact set is compact. Additionally, show that the intersection of a collection of closed sets containing a compact set or compact sets is also compact. #Answer# (a) To show that a closed subset F of a compact set K is compact, consider an open cover O of F. Extend the open cover by adding the complement of F in set A, and obtain a new open cover O'. Since K is compact, O' has a finite subcover O0' that covers K. Removing the complement of F from O0', we get the desired finite subcover O0 of F, proving that F is compact. (b) To show that the intersection F of a collection of closed sets containing a compact set ℋ is compact, consider an open cover O of F. Since ℋ is compact and F is a subset of ℋ, every open cover of F has a finite subcover for ℋ, say O0. As F is a subset of ℋ, this finite subcover O0 also covers F, proving that F is compact. (c) To show that the intersection F of a collection of compact sets is compact, consider an open cover O of F. For each compact set Ti in the collection, there exists a finite subcover Oi that covers Ti. Take the union of all these finite subcovers, and we get a finite subcover O0 for F, proving that F is compact.

Step-by-step solution

Consider an open cover of the closed subset

Let \(K\) be a compact set, and \(F\) be a closed subset of \(K\). Consider an open cover \(\mathcal{O} = \{O_\alpha\}\) of \(F\).

Extend the open cover

Add the open set \(A \setminus F\) (the complement of \(F\) in \(A\)) to the open cover \(\mathcal{O}\), obtaining a new open cover \(\mathcal{O}' = \{O_\alpha\} \cup \{A \setminus F\}\).

Use the compactness of K

Since \(K\) is compact, the open cover \(\mathcal{O}'\) has a finite subcover \(\mathcal{O}_0' \subseteq \mathcal{O}'\) that covers \(K\).

Remove the complement of F

Removing \(A \setminus F\) from the finite subcover \(\mathcal{O}_0'\), we get the desired finite subcover \(\mathcal{O}_0\) of \(F\). Therefore, \(F\) is compact. (b) Intersection of closed sets with a compact set is compact

Consider an open cover of the intersection

Let \(\mathcal{T}\) be a collection of closed sets with a compact set \(\widehat{T}\). Let \(F = \cap\\{T \mid T \in \mathcal{T}\\}\). Consider an open cover \(\mathcal{O}\) of \(F\).

Use the compactness of \(\widehat{T}\)

Since \(\widehat{T}\) is compact and \(F \subseteq \widehat{T}\), every open cover of \(F\) has a finite subcover for \(\widehat{T}\). Let \(\mathcal{O}_0 \subseteq \mathcal{O}\) be such a finite subcover for \(\widehat{T}\).

Show that \(\mathcal{O}_0\) covers F

Since \(F \subseteq \widehat{T}\), this finite subcover \(\mathcal{O}_0\) also covers \(F\). Therefore, \(F\) is compact. (c) Intersection of compact sets is compact

Consider an open cover of the intersection

Let \(\mathcal{T}\) be a collection of compact sets. Let \(F = \cap\\{T \mid T \in \mathcal{T}\\}\). Consider an open cover \(\mathcal{O}\) of \(F\).

For each compact set, find a finite subcover

For each compact set \(T_i \in \mathcal{T}\), there exists a finite subcover \(\mathcal{O}_i \subseteq \mathcal{O}\) such that \(T_i \subseteq \bigcup \mathcal{O}_i\).

Form a finite subcover for F

Take the union of all the finite subcovers \(\mathcal{O}_i\), and we get a finite subcover \(\mathcal{O}_0\) for \(F\). Therefore, \(F\) is compact.

Key concepts

Closed Sets

In mathematics, especially in topology, a closed set is a fundamental concept that helps determine whether certain operations or limits lie within a defined space. Closed sets have several interesting properties:

• A set is closed if its complement is open.
• Closed sets include their boundary points. This means, for example, intervals like \( [a, b] \) in the real number line are closed because they include the endpoints \( a \) and \( b \).
• The intersection of any collection of closed sets is also closed.

Let's relate this to compactness. Suppose we have a closed subset \( F \) of a compact set \( K \). The fact that \( F \) is closed means it contains its boundary, making it "contained" within the limits. When you consider an open cover, you can always find a finite subcover for \( K \), and by adjusting, also for \( F \), sustaining its compact nature.

Metric Space

A metric space is a set paired with a metric, a function that gives a distance between any two points in the set. This notion is crucial for spatial analysis in mathematics.

• A metric \( \rho(x, y) \) satisfies these conditions: non-negativity, \(\rho(x, y) \geq 0\) and \(\rho(x, y) = 0\) if and only if \(x = y\); symmetry, \(\rho(x, y) = \rho(y, x)\); and the triangle inequality, \(\rho(x, z) \leq \rho(x, y) + \rho(y, z)\).
• Examples include the Euclidean space \( \mathbb{R}^n \), where the distance is the familiar \"straight-line\" measure between points.

For closed sets within a metric space \( (A, \rho) \), compactness ensures that every open cover of a set has a finite subcover. Consider a compact subset \( \widehat{T} \) within a collection of closed sets in this space. When converging on intersections of such sets, you can still find finite covers, attributing compactness to their intersection.

Open Cover

An open cover is a collection of open sets that together cover a space. This means each point in the set lies within at least one of the open sets in the collection.

• Open sets are those sets that do not include their boundary points. For example, \( (a, b) \) is open because it doesn't contain \( a \) or \( b \).
• An open cover for a set \( F \) in space includes open sets \( \{O_\alpha\} \) such that \( F \subseteq \bigcup O_\alpha \).

Open covers relate deeply to compactness. A set is compact if every open cover has a finite subcover. This property is essential in simplifying the analysis of infinite situations. When dealing with collections of compact sets, like in a metric space, verifying compactness via these covers and their finite subcovers ensures the results maintain a certain robustness across intersections.