Question

Suppose that \(\mathbf{F}: \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}\) is continuous on a compact subset \(S\) of \(\mathbb{R}^{n}\). Show that \(\mathbf{F}(S)\) is a compact subset of \(\mathbb{R}^{m}\).

Short answer

Question: Prove that a continuous function maps compact sets to compact sets. Answer: To prove that a continuous function maps compact sets to compact sets, we need to show that for a continuous function F: ℝⁿ → ℝᵐ and a compact subset S ⊂ ℝⁿ, the image of the subset S under F, denoted as F(S), is also a compact subset of ℝᵐ. We achieve this by using the sequential definition of compactness, which states that a set is compact if every sequence in it has a converging subsequence with a limit in the set. Following a step by step solution, we first prepare the proof by considering an arbitrary sequence in F(S) and finding a corresponding sequence in S. Then, we show that the sequence in S has a convergent subsequence since S is compact. By using the continuity of the function F, we find the limit of the corresponding subsequence in F(S). We then show that the converging point is in F(S), demonstrating that F(S) is compact as every sequence drawn from it has a converging subsequence with a limit in F(S). Thus, the continuous function F maps compact sets to compact sets.

Step-by-step solution

Prepare the proof

To prove that \(\mathbf{F}(S)\) is compact, we will use the sequential definition of compactness. Let \(\{ \mathbf{x}_k \}_{k=1}^{\infty}\) be a sequence of points in \(\mathbf{F}(S)\). Since every point in this sequence is in the image of \(S\) under \(\mathbf{F}\), there exists a corresponding sequence \(\{ \mathbf{y}_k \}_{k=1}^{\infty}\) in \(S \subset \mathbb{R}^{n}\) such that \(\mathbf{F}(\mathbf{y}_k) = \mathbf{x}_k\) for each \(k\).

Show that the sequence in S has a convergent subsequence

Since \(S\) is compact, every sequence in \(S\) has a convergent subsequence. Thus, there exists a convergent subsequence \(\{ \mathbf{y}_{k_i} \}_{i=1}^{\infty}\) of \(\{ \mathbf{y}_k \}_{k=1}^{\infty}\), which converges to a point \(\mathbf{y}_{\infty}\) in \(S\). In other words, \(\lim_{i\to\infty} \mathbf{y}_{k_i} = \mathbf{y}_{\infty}\).

Use the continuity of F to get the limit of the corresponding subsequence in F(S)

Next, let's consider the subsequence \(\{ \mathbf{x}_{k_i} \}_{i=1}^{\infty}\) in \(\mathbf{F}(S)\), associated with the subsequence \(\{ \mathbf{y}_{k_i} \}_{i=1}^{\infty}\) in \(S\). Since \(\mathbf{F}\) is continuous, we have that $$ \lim_{i\to\infty} \mathbf{F}(\mathbf{y}_{k_i}) = \mathbf{F}(\lim_{i\to\infty} \mathbf{y}_{k_i}), $$ which means $$ \lim_{i\to\infty} \mathbf{x}_{k_i} = \mathbf{F}(\mathbf{y}_{\infty}). $$

Show that the converging point is in F(S)

Owing to the continuity of \(\mathbf{F}\), the limit point \(\mathbf{F}(\mathbf{y}_{\infty})\) must be in \(\mathbf{F}(S)\) since \(\mathbf{y}_{\infty} \in S\). Thus, we have found a convergent subsequence \(\{ \mathbf{x}_{k_i} \}_{i=1}^{\infty}\) of the sequence \(\{ \mathbf{x}_k \}_{k=1}^{\infty}\) in \(\mathbf{F}(S)\), which converges to a point in \(\mathbf{F}(S)\).

Conclude that F(S) is compact

Since the sequence \(\{ \mathbf{x}_k \}_{k=1}^{\infty}\) in \(\mathbf{F}(S)\) was arbitrary, we have shown that every sequence in \(\mathbf{F}(S)\) has a convergent subsequence with a limit in \(\mathbb{F}(S)\). Therefore, \(\mathbf{F}(S)\) is a compact subset of \(\mathbb{R}^{m}\).

Key concepts

Continuous Functions

In mathematics, a function is considered continuous if small changes in the input result in small changes in the output. This is a key concept because continuous functions preserve the limits of sequences. When dealing with a continuous function \(F: \mathbb{R}^{n} \rightarrow \\mathbb{R}^{m}\), it means that for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that whenever ||\(\mathbf{x} - \mathbf{y}|| < \delta\), then ||\(\mathbf{F}(\mathbf{x}) - \mathbf{F}(\mathbf{y})|| < \epsilon\).
When you apply a continuous function to a sequence with a limit, the function of that sequence also approaches the function of the limit. This property is essential in proving results related to compactness.

• Continuous functions maintain convergence of sequences.
• They are crucial in ensuring that the image of a compact set under a continuous function is also compact.

Compact Sets

A set is called compact if it is both closed and bounded. In the context of real analysis, compact sets in \(\mathbb{R}^n\) have special and useful properties.
Compactness is vital because it ensures that every sequence within the set has a limit point that also lies within the set. This kind of structure allows us to draw strong conclusions about sequences and function behavior on these sets.

• Closed sets contain all their limit points.
• Bounded sets have elements contained within a finite space.
• In \(\mathbb{R}^n\), a compact set offers predictability for sequence convergence.

This property is heavily employed in analysis to establish the compactness of the image of functions.

Sequential Compactness

The concept of sequential compactness focuses on sequences within a set. A set is sequentially compact if every sequence in the set has a subsequence that converges to a point in the same set.
This concept is very useful when working with continuous functions and compact sets. It is particularly employed in proofs dealing with limits and convergence. Sequential compactness is important as it often acts as a more accessible tool for proving compactness in real analysis.

• Sequential compactness simplifies the handling of limits and convergence.
• It helps to demonstrate compactness by verifying the convergence property of sequences.

In proving a function's image is compact, the property of sequential compactness is often employed to conclude that limits of sequences remain within the function's image set.

Subset Compactness

Subset compactness in real analysis refers to the compactness of a set based on the property that it is a subset of another compact set. If \(S\) is a compact subset of \(\mathbb{R}^n\), then any continuous image of \(S\) will also be compact in \(\mathbb{R}^m\).
This characteristic is an extension of compact sets and is key in various theorems, such as the one mentioned in the exercise. Through the function \(\mathbf{F}\)'s continuity and \(S\)'s compactness, the image \(\mathbf{F}(S)\) inherits compactness.

• This ensures predictability and control over function behaviors.
• It relies heavily on the notion of preserving compactness through transformations.

Convergence of Sequences

Convergence of sequences is a fundamental concept in analysis. A sequence \(\{x_n\}\) is said to converge to a limit \(L\) if for every \(\epsilon > 0\), there exists an integer \(N\) such that for all \(n > N\), the terms \(x_n\) are within \(\epsilon\) of \(L\).
Convergence tells us about the behavior of sequences and is a crucial concept in establishing sequential compactness. It plays a central role when understanding function behaviors such as continuity.

• Convergence maintains the trajectory of sequence values approaching a specific point.
• The property is critical in analysis proofs, helping to link continuous functions and compact sets.

In proving that \(\mathbf{F}(S)\) is compact, convergence underpins the argument by confirming the existence of subsequences within the set limits.