Question

Let \(f\) be a continuous real-valued function on a compact subset \(S\) of a metric space \((A, \rho) .\) Let \(\sigma\) be the usual metric on \(\mathbb{R} ;\) that is, \(\sigma(x, y)=|x-y| .\) (a) Show that \(f\) is bounded on \(S\). (b) Let \(\alpha=\inf _{u \in S} f(u)\) and \(\beta=\sup _{u \in S} f(u) .\) Show that there are points \(u_{1}\) and \(u_{2}\) in \([a, b]\) such that \(f\left(u_{1}\right)=\alpha\) and \(f\left(u_{2}\right)=\beta\).

Short answer

In this exercise, we have proved that a continuous real-valued function \(f\) defined on a compact subset \(S\) of a metric space \((A, \rho)\) is bounded on \(S\). Moreover, we have shown that there exist points \(u_1\) and \(u_2\) in \(S\) such that \(f(u_1) = \alpha\) (the infimum of \(f(u)\) for all \(u \in S\)) and \(f(u_2) = \beta\) (the supremum of \(f(u)\) for all \(u \in S\)). This result follows from the definitions of compactness and continuity, as well as the properties of infimum and supremum, and it demonstrates the importance of compactness and continuity in analysis.

Step-by-step solution

Understanding compactness and continuity

A set \(S\) is called compact if every open cover of \(S\) has a finite subcover. Informally, this means that we can cover the set \(S\) using finitely many open sets. The function \(f\) is said to be continuous on \(S\) if, given any point \(x \in S\) and any \(\epsilon > 0\), there exists a \(\delta > 0\) such that for all \(y \in S\) with \(\rho(x, y) < \delta\), we have \(\sigma(f(x), f(y)) < \epsilon\). Informally, this means that if \(x\) and \(y\) are close enough, then their images under \(f\) are also close.

Proving \(f\) is bounded on \(S\)

To prove that \(f\) is bounded, we need to show that there exists a constant \(M > 0\) such that for all \(u \in S\), we have \(|f(u)| \leq M\). Since \(f\) is continuous on the compact set \(S\), it is uniformly continuous. This means that, given any \(\epsilon > 0\), there exists a \(\delta > 0\) such that for all \(x, y \in S\) with \(\rho(x, y) < \delta\), we have \(\sigma(f(x), f(y)) < \epsilon\). Choose any \(\epsilon > 0\). For every \(u \in S\), consider the open ball \(B(u, \delta)\), centered at \(u\) with radius \(\delta\). These open balls cover the compact set \(S\). By compactness of \(S\), there exists a finite subcover of these open balls, say \({B(u_1, \delta), B(u_2, \delta), \cdots, B(u_n, \delta)}\). Now, consider $$M = \max_{1 \leq i \leq n} (|f(u_i)| + \epsilon).$$ We will show that for any \(u \in S\), we have \(|f(u)| \leq M\). Let \(u \in S\). Then, there exists some \(1 \leq j \leq n\) such that \(\rho(u, u_j) < \delta\). By the uniform continuity of \(f\), we have $$\sigma(f(u), f(u_j)) = |f(u) - f(u_j)| < \epsilon.$$ Hence, we have $$|f(u)| \leq |f(u_j)| + |f(u) - f(u_j)| < |f(u_j)| + \epsilon \leq M.$$ Therefore, \(f\) is bounded on \(S\).

Finding points \(u_1\) and \(u_2\) such that \(f(u_1) = \alpha\) and \(f(u_2) = \beta\)

We are given that \(\alpha = \inf_{u \in S} f(u)\) and \(\beta = \sup_{u \in S} f(u)\). Since \(f\) is continuous and bounded on the compact set \(S\), it attains its infimum and supremum on this set. This means that there exist points \(u_1\) and \(u_2\) in \(S\) such that \(f(u_1) = \alpha\) and \(f(u_2) = \beta\). To see why this is true, let \(A = \{f(u) \mid u \in S\}\). Since \(S\) is nonempty and \(f\) is continuous and bounded on \(S\), the set \(A\) is nonempty and bounded. Therefore, it has an infimum \(\alpha\) and a supremum \(\beta\). By the definition of infimum and supremum, there exist sequences \((x_n)\) and \((y_n)\) in \(S\) such that \(f(x_n) \to \alpha\) and \(f(y_n) \to \beta\) as \(n \to \infty\). Since \(S\) is compact, there exist convergent subsequences \((x_{n_k}) \to u_1\) and \((y_{n_k}) \to u_2\), where \(u_1, u_2 \in S\). Since \(f\) is continuous on \(S\), we have $$f(u_1) = \lim_{k \to \infty} f(x_{n_k}) = \alpha$$ and $$f(u_2) = \lim_{k \to \infty} f(y_{n_k}) = \beta.$$ This proves that there are points \(u_1\) and \(u_2\) in \(S\) such that \(f(u_1) = \alpha\) and \(f(u_2) = \beta\).

Key concepts

Continuous Functions

A continuous function is one that preserves the closeness of inputs to produce closeness of outputs. In the context of a metric space, consider a function \( f \) which maps elements from a set \( S \) into the real numbers. The function \( f \) is continuous if, for every point \( x \) in \( S \) and for every number \( \epsilon > 0 \), no matter how small, there exists another positive number \( \delta > 0 \) such that whenever two points \( x \) and \( y \) in \( S \) satisfy the distance \( \rho(x, y) < \delta \), the corresponding function values satisfy \( \sigma(f(x), f(y)) < \epsilon \).
This idea can be visualized as ensuring that close inputs (in terms of the metric \( \rho \)) lead to close outputs (in terms of the metric \( \sigma \)).
Continuous functions on compact spaces, such as the set \( S \) in our problem, have the important property of being uniformly continuous, meaning that the choice of \( \delta \) works uniformly over the entire set.

Bounded Functions

A function is bounded on a set if there is a real number \( M \) so that for every value \( f(u) \) in the range of the function over that set, \(|f(u)| \leq M\). Essentially, the outputs of the function do not "blow up" or become infinite.
For continuous functions defined on compact sets, like our subset \( S \) of the metric space, boundedness is automatically guaranteed due to the nature of compactness.

• Compactness implies that the set can be completely enclosed by a finite collection of open sets.
• This ensures a structured environment where functions can be examined consistently.
• Continuous functions on these compact sets therefore have a maximum and minimum value.

In essence, the idea of a function being bounded on a compact set removes any concerns about the function "running away" to infinity or negative infinity, keeping it "in check" because of the finite nature of compactness.

Metric Space

A metric space is a set equipped with a function, referred to as a **metric**, that defines a notion of distance between every pair of points in the set. The most familiar example of a metric is the Euclidean distance in \( \mathbb{R}^n \).
In our problem, we deal with a metric space \( (A, \rho) \), where \( \rho \) is the metric determining the distance between any two points in \( A \).

Key properties of a metric space include:

• **Positive Definiteness**: For all \( x, y \) in \( A \), \( \rho(x, y) \geq 0 \) with \( \rho(x, y) = 0 \) if and only if \( x = y \).
• **Symmetry**: For all \( x, y \) in \( A \), \( \rho(x, y) = \rho(y, x) \).
• **Triangle Inequality**: For all \( x, y, z \) in \( A \), \( \rho(x, z) \leq \rho(x, y) + \rho(y, z) \).

These properties allow us to rigorously talk about continuity and compactness within the metric space.
In the context of our problem, compactness of the subset \( S \) within the metric space \( A \) allows us to apply the compactness theorem, ensuring that continuous functions on \( S \) are not only bounded but also attain maximum and minimum values. This facilitates finding specific points that achieve the infimum \( \alpha \) and supremum \( \beta \) of the function's range.