Problem 4 - A Primer of Real Analysis

Question

Show that if \(K\) is compact and \(C \subset K\) is closed, then \(C\) is compact.

Step-by-Step Solution

Answer

If \(C\) is a closed subset of a compact set \(K\), then \(C\) is compact.

Step 1: Understand the Definitions

Recall that a set is compact if every open cover of the set has a finite subcover. A set is closed if it contains all its limit points.

Step 2: Set Up the Problem

Given that \(K\) is a compact set and \(C \subset K\) is a closed set, we need to show that \(C\) is also compact by verifying that every open cover of \(C\) has a finite subcover.

Step 3: Define an Open Cover of \(C\)

Let \( \{U_\alpha\}_{\alpha \in A} \) be an open cover for \(C\). This means \(C \subseteq \bigcup_{\alpha \in A} U_\alpha\) and each \(U_\alpha\) is open.

Step 4: Extend the Open Cover to \(K\)

Since \(C\) is closed in \(K\) and \(K\) is compact, consider also the open set \(V = K \setminus C\), which covers the remaining part of \(K\). Then \( \{U_\alpha\}_{\alpha \in A} \cup \{V\} \) forms an open cover of \(K\).

Step 5: Use the Compactness of \(K\)

Since \(K\) is compact, the open cover \( \{U_\alpha\}_{\alpha \in A} \cup \{V\} \) must have a finite subcover. This means there is a finite subcollection \( \{U_1, U_2, ..., U_n, V\} \) that covers \(K\).

Step 6: Conclude for \(C\)

Notice that the set \(V\) covers the part of \(K\) that is outside \(C\). Thus, \(C\) is covered by \( \{U_1, U_2, ..., U_n\} \), removing the need for \(V\) to cover \(C\). Hence, \( \{U_1, U_2, ..., U_n\} \) forms a finite subcover for \(C\).

Step 7: Conclude the Proof

Since every open cover of \(C\) has a finite subcover, \(C\) is compact.

Key Concepts

Open Cover

An open cover is a fundamental concept in topology, particularly in the context of compactness in real analysis. It consists of a collection of open sets whose union contains the set in question. Imagine painting a room where every wall must be covered entirely with paint.

• The paint represents the open sets in the open cover.
• The walls represent the set being covered.

For a set \( C \) to be compact, any way you choose the paint (the open sets), the room (set \( C \)) should still be fully covered with some finite selection of your paint cans.In the context of our exercise, \(\{U_\alpha\}_{\alpha \in A}\) is an open cover for the set \(C\). This means each element in \( C \) is contained within one or more of these open sets. Therefore, understanding open covers is key to proving whether a subset like \( C \) is compact, as it ensures all elements of the set can be covered finitely.

Closed Set

Closed sets are strikingly different from open sets. A closed set includes its boundary or limit points. Visualize it as a jar with a lid that stores not only its contents but also anything that tries to escape.

• Closed sets include all their edge points—imagine if the boundary was glued to the main body.
• This characteristic helps in understanding convergence and limit points within topology.

In the problem, \(C\) is closed in \(K\), which inherently means it includes its limit points and those points can't 'flux out'. To demonstrate that \(C\) is compact, we leveraged the closed set property, which facilitated creating a finite subcover from the open cover enveloping \( C \) and the rest of \( K \). Hence, closed sets possess a confinement quality essential when discussing compactness in mathematics.

Subcover

A subcover is derived from an open cover and retains its potency in providing coverage for a set. Consider a subcover as a detailed map representing only the lines and regions necessary to navigate a particular area without unnecessary complexity.

• A finite subcover means you can cover the original set with a finite number of open sets from the open cover.
• Finding a finite subcover is crucial in demonstrating compactness.

In our exercise, \(\{U_1, U_2, ..., U_n, V\}\) forms a finite subcover for \(K\), ensuring all of \(K\) is covered effectively. However, because \(V\) addresses the area outside \(C\), the finite subcover of \(C\) can solely ignore \(V\), focusing only on \(\{U_1, U_2, ..., U_n\}\). Accordingly, this distinction illustrates how subcovers work in fine-tuning the coverage needed to delineate a specific set's compactness.