Problem 9 - Basic Analysis: Introduction to Real Analysis

Question

Let \((X, d)\) be a nonempty metric space and \(S \subset X\) a subset. Prove: a) \(S\) is bounded if and only if for every \(p \in X,\) there exists \(a B>0\) such that \(d(p, x) \leq B\) for all \(x \in S\). b) A nonempty \(S\) is bounded if and only if \(\operatorname{diam}(S):=\sup \\{d(x, y): x, y \in S\\}<\infty\).

Step-by-Step Solution

Answer

For (a), S is bounded if distances from any point in X to S are bounded. For (b), S is bounded if its diameter is finite.

Step 1: Define Bounded Set

A subset \( S \) of a metric space \( (X, d) \) is said to be bounded if there exists some \( M > 0 \) such that \( d(x, y) \leq M \) for all \( x, y \in S \). This is equivalent to saying that the entire subset can fit within a ball of finite radius.

Step 2: Prove Part (a) Forward Direction

Assume \( S \) is bounded. For any point \( p \in X \), let \( M \) be such that \( d(s, t) \leq M \) for all \( s, t \in S \). Then for any \( x \in S \), choosing \( t \in S \) and noting \( S \subset X \), we have \( d(p, x) \leq d(p, t) + d(t, x) \leq d(p, t) + M \). Therefore, setting \( B = M + d(p, t) \) ensures the inequality holds for every \( x \in S \).

Step 3: Prove Part (a) Backward Direction

Assume that for every \( p \in X \), there is a \( B > 0 \) such that \( d(p, x) \leq B \) for all \( x \in S \). Pick any pair \( x, y \in S \), and apply the triangle inequality: \( d(x, y) \leq d(x, p) + d(p, y) \leq B + B = 2B \). Thus, \( S \) is bounded.

Step 4: Understand Diameter as a Boundedness Measure

The diameter of a set \( S \), denoted as \( \operatorname{diam}(S) \), is defined as \( \operatorname{diam}(S) = \sup \{ d(x, y) : x, y \in S \} \). This represents the maximum distance between any two points in \( S \). If this maximum distance is finite, \( S \) is bounded.

Step 5: Prove Part (b) Forward Direction

Suppose \( S \) is bounded, meaning there exists \( M > 0 \) such that \( d(x, y) \leq M \) for all \( x, y \in S \). Then by definition, \( \operatorname{diam}(S) \leq M \), which is finite. This shows boundedness implies finite diameter.

Step 6: Prove Part (b) Backward Direction

If \( \operatorname{diam}(S) < \infty \), there is an upper bound \( M = \operatorname{diam}(S) \) such that for all \( x, y \in S \), \( d(x, y) \leq M \). This finite \( M \) satisfies the definition of a bounded set. Therefore, set \( S \) is bounded by \( M \).

Key Concepts

Bounded Set

In the context of metric spaces, a "bounded set" refers to a subset where all points are within a certain distance from each other. Imagine all points being enclosed within a ball of finite radius in the space. This means if you pick any two points within the set, the distance between them will not exceed some fixed value. 
To be more precise, a subset \( S \) of a metric space \( (X, d) \) is considered bounded if there exists a number \( M > 0 \) such that for all points \( x, y \in S \), the distance \( d(x, y) \) is less than or equal to \( M \).
It's like saying, "No point in this set is further than \( M \) units from another." This concept is crucial because it helps us determine how "spread out" a set is within a space.

Triangle Inequality

The triangle inequality is a fundamental property of metric spaces that helps us understand distances in a more intuitive way. If you think of three points forming a triangle, this inequality states that the direct path between two points is always the shortest.
The mathematical expression of the triangle inequality is: for any points \( x, y, z \) in a metric space, the distance \( d(x, z) \) must satisfy:
\[ d(x, z) \leq d(x, y) + d(y, z) \].
This means the distance from \( x \) to \( z \) is never greater than the sum of distances from \( x \) to \( y \) and from \( y \) to \( z \), which essentially means taking a detour can't be shorter. This property is quite handy when proving the boundedness of sets because it allows us to relate different distances effectively.

Diameter of a Set

The "diameter of a set" is a measure that provides the longest possible distance between any two points within a set. Think of it as stretching the set to its fullest extent. For any subset \( S \) of a metric space, the diameter is calculated using the formula:
\[ \operatorname{diam}(S) = \sup \{ d(x, y) : x, y \in S \} \].
Here, the "\( \sup \)" refers to the supremum, or the least upper bound, for the set of distances between points in \( S \).
This concept helps us understand if a set is bounded—if the diameter is finite, we can say the set is bounded. Essentially, if there's a maximum distance that any two points in the set cannot exceed, the set can be "fit" into a finite space, confirming its boundedness.