Problem 4 - Advanced Calculus

Question

If \(A \subset B\) and \(B\) is a bounded set in \(n\) space, show that diam \((A) \leq \operatorname{diam}(B)\). Can equit lit? occur without having \(A=B\) ?

Step-by-Step Solution

Answer

Yes, the diameter of set \(A\) would always be less than or equal to the diameter of set \(B\). Equality can occur without \(A\) being equal to \(B\), for instance when \(A = [0, 1)\) and \(B = [0, 1]\).

Step 1: Understanding Definition of Diameter

The diameter of a set \(A\), \(diam(A)\), is the supremum of the set of distances \(\{d(x, y): x, y \in A\}\). Note that the supremum of a set is the least upper bound.

Step 2: Demonstrating Inequality

Since \(A\) is a subset of \(B\), all distances between pairs of points in \(A\) will be less than or equal to the corresponding distances in \(B\). Therefore, \(diam(A) \leq diam(B)\).

Step 3: Checking possibility of equality

Equality, \(diam(A) = diam(B)\), can occur without \(A\) being equal to \(B\). Consider a bounded set \(B = [0, 1]\) and a subset \(A = [0, 1)\). The diameter of both sets is 1 (distance between 0 and 1), but the sets are not equal.

Key Concepts

Supremum of a Set

In mathematics, the concept of the supremum, or 'least upper bound', is fundamental when discussing the sizes of sets. For a set of real numbers or in a more general context, a set in a metric space, the supremum is the smallest value that is greater than or equal to every element in the set. It's particularly important when considering distances, as in the diameter of a set.

The diameter of a set in a given space is a measure of how 'wide' the set is. Concretely, it is the supremum of the distances between all possible pairs of points within the set. This can be expressed mathematically as
\[ \operatorname{diam}(A) = \sup\{d(x, y) : x, y \in A\} \],where \( d(x, y) \) is a function defining the distance between two points \( x \) and \( y \) within the set \( A \). Understanding the diameter requires the comprehension of the supremum, as it relies on finding the upper bound of all possible distances. This bound must exist for a set to be considered bounded, another concept integral to the exercise.

Subset and Bounded Sets

A deeper dive into the problem involves two additional concepts: subsets and bounded sets. A set \( A \) is considered a subset of another set \( B \) if every element of \( A \) is also an element of \( B \); mathematically, it's expressed as \( A \subset B \).

Now let's consider boundedness. A set in n-dimensional space is considered bounded if there is a real number that serves as an upper limit to the distance between any two points in the set. In simpler terms, you can think of a bounded set as one that could be 'contained' within a finite n-dimensional sphere of some radius. This makes the concept of the diameter quite inherent to bounded sets as it inherently provides that limiting distance.

The exercise poses an interesting scenario: whether a subset \( A \) of a bounded set \( B \) can have the same diameter without being the exact same set. The subset, being entirely contained within \( B \), cannot have a larger diameter by definition, but it can indeed have the same diameter, as long as the 'extremes' or 'bounds' of \( A \) are the same as those of \( B \), without \( A \) necessarily including all points that \( B \) contains.

Metric Space Properties

Finally, it's essential to understand the properties of a metric space to fully grasp the context of the exercise. A metric space is a set equipped with a function called a 'metric' that defines the distance between any two points within the set. The metric has to follow a specific set of properties; it must be positive, symmetric, and obey the triangle inequality.

Some simple properties of a metric that are pertinent include:

• Distance is always non-negative: \( d(x, y) \geq 0 \).
• Distance is zero if and only if the two points are the same: \( d(x, y) = 0 \) iff \( x = y \).
• The distance from point \( x \) to point \( y \) is the same as from \( y \) to \( x \): \( d(x, y) = d(y, x) \).
• The triangle inequality holds: for any three points \( x, y, \) and \( z \), the distance from \( x \) to \( z \) is less than or equal to the sum of the distances from \( x \) to \( y \) and \( y \) to \( z \): \( d(x, z) \leq d(x, y) + d(y, z) \).

When talking about diameters within metric spaces, these properties ensure that the concept of distance remains consistent and meaningful. In the context of the exercise, the set \( A \) and its superset \( B \) reside in such a space, and their diameters are defined in terms of the distances given by the space's metric.