Chapter 5: Problem 20
Prove that \(d(S)=d(\bar{S})\) for any set \(S\) in \(\mathbb{R}^{n}\).
Short Answer
Expert verified
Answer: Yes, the diameter of a set is equal to the diameter of its closure in a metric space. This can be represented mathematically as \(d(S) = d(\bar{S})\).
Step by step solution
01
Define the diameter of a set
The diameter of a set \(S\) in a metric space is defined as the largest distance between any two points in the set. Mathematically, it is given by:
$$d(S) = \sup\{d(x, y) : x, y \in S\}$$
where \(d(x, y)\) represents the distance between points \(x\) and \(y\).
02
Define the closure of a set
The closure \(\bar{S}\) of a set \(S\) in a topological space is the smallest closed set containing \(S\). It is the union of the set \(S\) and its limit points.
03
Prove \(d(S) \leq d(\bar{S})\)
We need to show that the diameter of the set \(S\) is less than or equal to the diameter of its closure. Since \(S \subseteq \bar{S}\), the distances between any two points in \(S\) are also distances between points in \(\bar{S}\). Therefore, the supremum of the distances in \(S\) cannot exceed the supremum of the distances in \(\bar{S}\). Thus, we have
$$d(S) \leq d(\bar{S})$$
04
Prove \(d(\bar{S}) \leq d(S)\)
We need to show that the diameter of the closure of \(S\) is less than or equal to the diameter of \(S\). Let \(x, y \in \bar{S}\). By the definition of closure, there exist sequences \((x_n)\) and \((y_n)\) in \(S\) such that \(x_n \rightarrow x\) and \(y_n \rightarrow y\) as n approaches infinity.
Using the triangle inequality, we get:
$$d(x, y) \leq d(x, x_n) + d(x_n, y_n) + d(y_n, y)$$
Taking the limit as \(n \rightarrow \infty\), we have:
$$d(x, y) \leq \lim_{n\rightarrow\infty} (d(x, x_n) + d(x_n, y_n) + d(y_n, y))$$
Since \(x_n \rightarrow x\) and \(y_n \rightarrow y\), we get:
$$\lim_{n\rightarrow\infty} d(x, x_n) = 0$$
and
$$\lim_{n\rightarrow\infty} d(y, y_n) = 0$$
Therefore,
$$d(x, y) \leq \lim_{n\rightarrow\infty} d(x_n, y_n) \leq d(S)$$
As this holds for any \(x, y \in \bar{S}\), we have:
$$d(\bar{S}) \leq d(S)$$
05
Conclude the proof
Since we have shown that \(d(S) \leq d(\bar{S})\) and \(d(\bar{S}) \leq d(S)\), we can now conclude the proof by stating that the diameters are equal:
$$d(S) = d(\bar{S})$$
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Closure of a Set
In the realm of metric spaces, understanding the closure of a set is paramount. The closure of a set, denoted as \( \bar{S} \), refers to the smallest closed set that contains the original set \( S \). A closed set in this context is one that includes all its limit points—points where sequences in \( S \) converge to.
To visualize this, imagine \( S \) as a swarm of points scattered within a boundary. The closure encompasses all these points plus any outer edge or limit points reached by sequences originating from \( S \). This ensures that \( \bar{S} \) is robust, capturing every possible pathway these sequences can take.
For example, in real numbers, the closure of an open interval \((a, b) \) is the closed interval \([a, b]\), which includes all the values between \(a\) and \(b\) as well as the endpoints themselves, securing no gaps at the boundaries.
To visualize this, imagine \( S \) as a swarm of points scattered within a boundary. The closure encompasses all these points plus any outer edge or limit points reached by sequences originating from \( S \). This ensures that \( \bar{S} \) is robust, capturing every possible pathway these sequences can take.
For example, in real numbers, the closure of an open interval \((a, b) \) is the closed interval \([a, b]\), which includes all the values between \(a\) and \(b\) as well as the endpoints themselves, securing no gaps at the boundaries.
Limit Points
A core component of a set's closure, limit points add depth to understanding sequence behaviors within metric spaces. A limit point of a set \( S \) is a point that can be approached arbitrarily closely by elements within \( S \) itself. This means for any chosen proximity, or epsilon, there exist points in \( S \) within that distance from the limit point.
- For instance, consider the set of rational numbers within the interval \((0, 1)\). The irrational numbers, although not inside \( S \), are limit points, as subsets of the rationals can get infinitely close to any irrational number.
- Another example lies in sequences. If a sequence of points within a set converges to a particular point, that point serves as a limit point of the set.
Triangle Inequality
The triangle inequality is a fundamental property in metric spaces, pivotal for proving relationships like diameters in this exercise. This principle states that for any three points \( x, y, \) and \( z \) in a metric space, the direct distance between two points is always less than or equal to the distance traveled via an intermediary point. Formally, it is expressed as:
\[d(x, y) \leq d(x, z) + d(z, y)\]This rule resembles the geometric property seen in triangles: any side length must be less than or equal to the sum of the other two sides.
\[d(x, y) \leq d(x, z) + d(z, y)\]This rule resembles the geometric property seen in triangles: any side length must be less than or equal to the sum of the other two sides.
- In the context of this problem, when \( x \) and \( y \) are points within the closure \( \bar{S} \), sequences \( x_n \) and \( y_n \) drawn from \( S \) converge to \( x \) and \( y \). Here, the triangle inequality helps bridge the direct distance between these converging sequences and their limit points.
- This makes it a cornerstone in affirming the equality between the diameter of \( S \) and its closure.