Question

Show that a subset of a finite-dimensional normed linear space is compact if and only if it is closed and bounded. The next four results take the subject a little further. Some of them will form a key part of our later discussions of the inverse and implicit function theorems, but each is interesting in its own right, and provides experience of working in the space \(\mathcal{B}(X, Y)\)

Short answer

Question: Prove that a subset of a finite-dimensional normed linear space is compact if and only if it is closed and bounded. Solution: To prove this, we need to show two directions: 1. If a subset of a finite-dimensional normed linear space is compact, then it is closed and bounded. 2. If a subset of a finite-dimensional normed linear space is closed and bounded, then it is compact. For the first direction, we assume that a subset A is compact and show that A is closed and bounded using the definition of compactness and the properties of closed and bounded sets. For the second direction, we assume that A is closed and bounded and show that A is compact using the Heine-Borel theorem and the properties of the finite-dimensional normed linear space. Together, these two directions establish the equivalence between compactness and being closed and bounded in a finite-dimensional normed linear space.

Step-by-step solution

First direction: From compact to closed and bounded

Assume that a subset \(A\) of a finite-dimensional normed linear space \(V\) is compact. We want to show that \(A\) is closed and bounded.

Show that A is closed

We want to show that the complement of \(A\) in \(V\), denoted by \(A^c\), is open. If \(x \in A^c\), then for every \(y \in A\), we have that the distance between \(x\) and \(y\), \(\|x - y\| > 0\). Since \(A\) is compact, there exists a finite open cover of \(A\), say \(\{B(y_i, r_i)\}_{i=1}^n\) where \(B(y_i, r_i)\) is an open ball centered at \(y_i\) with radius \(r_i\). Let \(r = \min\{r_1,\ldots,r_n\}\). Define an open ball \(B(x, r)\) centered at \(x\) with radius \(r\). Clearly, \(B(x, r) \subset A^c\), and so \(A^c\) is open, which implies that \(A\) is closed.

Show that A is bounded

Since \(A\) is compact, it is covered by a finite collection of open balls \(\{B(y_i, r_i)\}_{i=1}^n\). Let \(R = \max\{\|y_1\| + r_1, \ldots, \|y_n\| + r_n\}\). Then, for every \(x \in A\), there exists an index \(i\) such that \(x \in B(y_i, r_i)\). Thus, we have \(\|x\| \leq \|y_i\| + r_i \leq R\). Therefore, \(A\) is bounded. Now we have proven the first direction. Let's proceed with the other direction.

Second direction: From closed and bounded to compact

Assume that a subset \(A\) of a finite-dimensional normed linear space \(V\) is both closed and bounded. We want to show that \(A\) is compact.

Apply Heine-Borel theorem

Since \(V\) is a finite-dimensional normed linear space, it is isometrically isomorphic to a Euclidean space \(\mathbb{R}^n\) for some \(n\). Let \(\phi : V \to \mathbb{R}^n\) be an isomorphism. Define \(\tilde{A} = \phi(A) \subseteq \mathbb{R}^n\). Since \(A\) is closed, and \(\phi\) is a homeomorphism, \(\tilde{A}\) is also closed. Since \(A\) is bounded, there exists an \(R > 0\) such that \(\|x\| \leq R\) for every \(x \in A\). Then, \(\|\phi(x)\| \leq R\) for every \(x \in A\), which implies that \(\tilde{A}\) is bounded. Now, apply the Heine-Borel theorem to the closed and bounded set \(\tilde{A} \subseteq \mathbb{R}^n\), which states that \(\tilde{A}\) is compact.

Show that A is compact

We want to show that every open cover of \(A\) has a finite subcover. Let \(\{G_{\alpha}\}\) be an open cover of \(A\). Then, \(\{\phi^{-1}(G_{\alpha})\}\) is an open cover of \(\tilde{A}\). Since \(\tilde{A}\) is compact, there exists a finite subcover \(\{\phi^{-1}(G_{\alpha_1}), \ldots, \phi^{-1}(G_{\alpha_m})\}\). Then, \(\{G_{\alpha_1},\ldots,G_{\alpha_m}\}\) is a finite subcover of \(A\), which implies that \(A\) is compact. Thus, we have shown that if a subset of a finite-dimensional normed linear space is closed and bounded, then it is compact, completing the proof.

Key concepts

Finite-dimensional normed linear space

A finite-dimensional normed linear space is a vector space equipped with a norm, where the dimension of the space is finite. This means that the space can be spanned by a finite number of vectors. Having a norm allows us to measure the size or length of vectors in the space. This is crucial for defining concepts like distance and convergence. In simpler terms, think of it as a geometric space with a limited number of directions.

Finite-dimensional spaces are significant because they behave very much like the familiar Euclidean spaces, such as \(\mathbb{R}^n\). This similarity allows many mathematical properties and theorems applicable to Euclidean spaces to be used in finite-dimensional normed spaces, simplifying analysis and problem-solving in such environments.

Heine-Borel theorem

The Heine-Borel theorem is a fundamental result in real analysis that applies to Euclidean spaces. It states that a subset of \(\mathbb{R}^n\) is compact if and only if it is closed and bounded. This theorem helps us identify compact sets easily, using the properties of being closed and bounded.

In the context of finite-dimensional normed linear spaces, we can use an isomorphism to treat these spaces as Euclidean spaces. Thus, the Heine-Borel theorem becomes applicable here too, allowing us to deduce that sets that are both closed and bounded are compact.

Compactness is vital because it implies that every sequence within the set has a subsequence that converges to a point in the set, making analysis and calculations more manageable.

Closed and bounded sets

In mathematical analysis, a set is closed if it contains all its limit points. This means you can't "escape" the set by approaching a boundary point that isn't included in the set itself. A set is bounded if there exists a real number that serves as a limit to the "size" of the set.

To put it simply:

• A closed set includes its edges, like a fence that includes the land it surrounds.
• A bounded set has a limit to how far it stretches in any direction.

When a set in a finite-dimensional normed linear space is both closed and bounded, we can invoke the Heine-Borel theorem to conclude that the set is compact, which significantly aids in the study of functions and sequence behaviors within the set.

Isomorphic to Euclidean space

An isomorphism between two mathematical structures is a mapping that shows a one-to-one correspondence between them. When a finite-dimensional normed linear space is isomorphic to a Euclidean space, it implies they share the same structural properties.

This concept is powerful because it lets us transfer intuitive geometric concepts from Euclidean spaces to abstract finite-dimensional normed spaces. The isomorphism ensures that properties like distances and angles are preserved, making it easier to apply familiar mathematical theorems and techniques.

Open and closed sets

Open and closed sets are fundamental concepts in topology and analysis. An open set is one where, for any point within the set, there is some room to move around in any direction without leaving the set. This "wiggle room" is defined by the existence of an open ball centered at that point.

A closed set, on the other hand, includes all its boundary points. If you think of a closed set like a sealed envelope, everything inside, including the edges, is part of the set.

Together, these concepts help us understand continuity, convergence, and compactness in mathematical spaces. For compactness, knowing that sets are both closed and bounded is essential in finite-dimensional spaces, allowing us to apply the Heine-Borel theorem effectively.