Problem 12

Question

Let $(X, d)$ be a complete metric space. Show that $K \subset X$ is compact if and only if $K$ is closed and such that for every $\varepsilon>0$ there exists a finite set of points $x_{1}, x_{2}, \ldots, x_{n}$ with $K \subset \bigcup_{j=1}^{n} B\left(x_{j}, \varepsilon\right) .$ Note: Such a set $K$ is said to be totally bounded, so in a complete metric space a set is compact if and only if it is closed and totally bounded.

Step-by-Step Solution

Verified

Answer

A set in a complete metric space is compact if and only if it is closed and totally bounded.

1Step 1: Understanding the Problem

We need to show that in a complete metric space, a subset $ K $ is compact if and only if it is closed and totally bounded. A set is totally bounded if for every $ \varepsilon > 0 $ there exists a finite number of points whose $ \varepsilon $-balls cover $ K $.

2Step 2: Implication 1: Compact implies Closed and Totally Bounded

First, assume $ K $ is compact. A compact set in a metric space is closed and bounded. A bounded set is totally bounded in a complete metric space because we can cover it with a finite number of $ \varepsilon $-balls, thereby establishing it is totally bounded.

3Step 3: Implication 2: Closed and Totally Bounded implies Compact

Now, assume $ K $ is closed and totally bounded. A complete metric space implies every Cauchy sequence in $ K $ converges to a point in $ K $. Totally bounded means $ K $ can be covered with a finite number of $ \varepsilon $-balls for any $ \varepsilon > 0 $, which ensures any sequence in $ K $ can be contained within a compact subset concluding $ K $ is compact by Bolzano-Weierstrass theorem.

4Step 4: Conclude with Metric Space Properties

In a complete metric space, the properties of being closed and totally bounded indeed suffice to ensure compactness by combining sequence compactness and covering properties. Thus, through the implications studied, we've shown that a set is compact if and only if it's closed and totally bounded.

Key Concepts

Complete Metric SpaceTotally Bounded SetsClosed SetsBolzano-Weierstrass Theorem

Complete Metric Space

A complete metric space is a very special type of space in metric topology. It has a unique property where every Cauchy sequence has a limit that is also within the space. This idea is significant because it ensures that sequences behave nicely without jumping out of the given space. Here’s what you should know about complete metric spaces:

Cauchy sequences: In these sequences, the terms become arbitrarily close to each other as the sequence progresses.
Convergence within the space: Complete metric spaces guarantee that if a sequence converges, its limit will also be a point within the space itself, ensuring closure in a specific sense.

Understanding complete metric spaces is crucial because the compactness properties we're discussing about depend heavily on the completeness of the space.

Totally Bounded Sets

Totally bounded sets are an interesting concept that is a bit stronger than simply being bounded. A set is said to be totally bounded if for any given positive distance, known as $ \varepsilon $, you can cover the entire set using a finite number of $ \varepsilon $-balls.
Think of it this way:

A set is covered by small overlapping circles, where each has a radius of $ \varepsilon $. These contribute to a finite cover, making the set totally bounded.
Totally bounded is a precursor for compactness in a complete metric space, alongside being closed.

Totally bounded sets hint towards a property of controlling how spread out a set can be, ensuring that no matter how you zoom in, you can cover it with a finite number of small regions.

Closed Sets

In a topological space, closed sets have a specific, neat boundary idea. A set is closed if it contains all its limit points. This means if you have a sequence of points within the set that converge to a certain limit, that limit must also lie within the set.
Let's explore what makes a set closed:

Boundary inclusion: A set that includes its edge points is closed. If you reach the boundary, you're still inside the set.
Complementary Open Concept: A set is closed if its complement (everything outside the set) is open.

Closed sets are essential in linking the concept of compactness with complete metric spaces, as they ensure no sequence can "escape" the set. In our context, a set being closed fits perfectly with being totally bounded to become compact.

Bolzano-Weierstrass Theorem

The Bolzano-Weierstrass Theorem is a key theorem in analysis and relates to sequences within spaces. It is the go-to tool when discussing compactness in metric spaces.
Here’s a simple breakdown:

Converging subsequences: The theorem states that in any bounded sequence in $ \mathbb{R}^n $, there exists a convergent subsequence.
Linking to compactness: This theorem helps show that every bounded and closed set in $ \mathbb{R}^n $ is compact, as it ensures sequences cannot "wander off" endlessly without converging.

In the context of our original problem, the Bolzano-Weierstrass theorem underpins the proof that a totally bounded and closed set is compact. It guarantees sequence convergence, completing the path from a bounded, closed set to a compact set in a complete metric space.

Problem 12

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