Question

Let \(X\) be a metric space in which every infinite subset has a limit point. Prove that \(X\) is compact. Hint: By Exercises 23 and \(24, X\) has a countable base. It follows that every open cover of \(X\) has a countable subcover \(\left\\{G_{a}\right\\}, n=1,2,3, \ldots\) If no finite subcollection of \(\left\\{G_{n}\right\\}\) covers \(X\), then the complement \(F_{n}\) of \(G_{1} \cup \cdots \cup G_{n}\) is nonempty for each \(n\), but \(\cap F_{n}\) is empty. If \(E\) is a set which contains a point from each \(F_{n}\), consider a limit point of \(E\), and obtain a contradiction.

Short answer

Question: Prove that a metric space X is compact if every infinite subset of X has a limit point. Answer: To prove that a metric space X is compact given every infinite subset of X has a limit point, we can follow the steps outlined in the solution: 1. Show that X has a countable base. 2. Assume no finite subcollection of the countable open cover covers X. 3. Define the complement F_n. 4. Consider the intersection of all F_n. 5. Construct the set E. 6. Find a limit point of E. 7. Prove a contradiction by considering the set G_i containing infinitely many points of E. 8. Conclude that X is indeed compact, as there must exist a finite subcollection of the countable open cover that covers X.

Step-by-step solution

Show that X has a countable base

By Exercises 23 and 24, we know that X has a countable base. Let this be the collection of open sets \(\{B_n\}_{n=1}^{\infty}\).

Assume no finite subcollection of open cover covers X

Let \(\{G_n\}_{n=1}^{\infty}\) be a countable open cover of X such that no finite subcollection of \(\{G_n\}\) covers X.

Define the complement F_n

Define \(F_n\) as the complement of the union of the first n sets of the countable open cover, i.e., \(F_n = X - (G_1 \cup G_2 \cup \cdots \cup G_n)\). Since no finite subcollection of \(\{G_n\}\) covers X, each \(F_n\) must be nonempty.

Consider the intersection of all F_n

Since each \(F_n\) is the complement of an increasing union of open sets, \(\{F_n\}\) forms a decreasing sequence of closed sets. Additionally, since no finite subcollection of \(\{G_n\}\) covers X, \(\bigcap_{n=1}^{\infty} F_n = \emptyset\).

Construct the set E

Construct a set E containing exactly one point from each \(F_n\). Since we know that each \(F_n\) is nonempty, E must be infinite.

Find a limit point of E

Since every infinite subset of X has a limit point, E must have a limit point as well. Let this limit point be x.

Prove a contradiction

Since \(\{G_n\}_{n=1}^{\infty}\) is an open cover, there exists some open set \(G_i\) containing x. By definition, since x is a limit point of E, there must exist an open set \(B_n \subset G_i\) containing x and infinitely many points of E. However, for all \(n \geq i\), E cannot contain any point of \(F_n\), since \(F_n \cap G_i = \emptyset\). The fact that there is a set \(G_i\) containing infinitely many points of E contradicts our construction that the set E should contain one point from each \(F_n\). Thus, the assumption that no finite subcollection of open cover covers X leads to a contradiction.

Conclusion

Since the assumption leads to a contradiction, it implies that there must exist a finite subcollection of the countable open cover \(\{G_n\}_{n=1}^{\infty}\) that covers X, meaning that X is indeed compact.

Key concepts

Limit Points in Metric Spaces

In metric spaces, understanding limit points is crucial. A limit point of a set is a point that can be approached by other points of the set. If we have a set \( S \) in a metric space \( X \), a point \( x \) is a limit point of \( S \) if every neighborhood of \( x \) contains at least one point of \( S \) different from \( x \) itself. This means you can get arbitrarily close to \( x \) using points from \( S \).

In our context, an important property given is that every infinite subset of \( X \) has a limit point. This ensures that we can't separate points in a way that avoids limit points, which is a key component of compactness.

Having limit points makes it difficult for sequences or sets to "escape" the space, hinting at the idea of compactness, where no infinite set can lie entirely outside a finite cover.

Countable Base in Topology

The concept of a countable base is foundational in topology. A topological space \( X \) is said to have a countable base if we can describe its topology through a countable collection of open sets. This collection, called a basis, can be used to generate all the open sets in the space by taking unions.

For example, the open intervals with rational endpoints form a countable base for the real numbers with the standard topology. In our exercise, we know \( X \) has a countable base, which is significant for proving compactness.

Why is a countable base important? It simplifies the complexity of the space by ensuring topological operations can be conducted within a framework of countable sets. This aids in forming countable subcovers, a stepping stone to proving compactness.

Open Cover and Subcover

An open cover of a space \( X \) is a collection of open sets that collectively cover \( X \); every element of \( X \) is in at least one of these open sets. Formally, if \( \{U_\alpha\} \) is an open cover of \( X \), then \( X = \bigcup_\alpha U_\alpha \).

A subcover is a smaller collection of sets from the original open cover that still covers \( X \). Finding a subcover can be crucial in compactness proofs to show that you don't need an infinite number of sets to cover the space.

The exercise involves showing that any open cover of \( X \) has a finite subcover. First, since \( X \) has a countable base, any open cover has a countable subcover. Then, using the property of finite intersection from the next section, we can extract a finite subcover.

Finite Subcollection and Intersection

The concept of finite subcollections is vital in understanding compactness. Given a collection of open sets covering a space \( X \), a finite subcollection covers \( X \) if the intersection of the complements outside the finite subcollection is empty.

This is connected to the exercise where \( F_n = X - (G_1 \cup G_2 \cup \cdots \cup G_n) \) for an open cover \( \{G_n\} \). If no finite subcollection covers \( X \), the intersection of all \( F_n \) should be non-empty. However, proving it empty under the correct conditions leads to a contradiction that enables proving compactness.

This concept checks that if you assume the space isn't covered by any finite subcollection, you're led to an impossibility, hinting that such a finite subcollection must exist, underpinning compactness.