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A set \(A\) in \((S, \rho)\) is said to be totally bounded iff for every \(\varepsilon>0\) (no matter how small \(), A\) is contained in a finite union of globes of radius \varepsilon. By Problem 3 , any such set is bounded. Disprove the converse by a counterexample. [Hint: Take an infinite set in a discrete space.]

Short Answer

Expert verified
The infinite discrete set \( \mathbb{N} \) is bounded but not totally bounded.

Step by step solution

01

Understanding the Definitions

Firstly, we need to understand the terms *totally bounded* and *bounded*. A set A in a metric space (S, ρ) is totally bounded if for every \( \varepsilon > 0 \), \( A \) can be covered by a finite number of open balls (or globes) of radius \( \varepsilon \). A set is bounded if it can be contained within some fixed distance from a point.
02

Identifying a Suitable Metric Space

The hint suggests using a discrete space. In a discrete metric space, the distance between any two distinct points is 1, and the distance from a point to itself is 0. This simplifies how distance functions in this space.
03

Selecting an Infinite Set in the Discrete Space

Consider an infinite set \( A \) that consists of all natural numbers \( \mathbb{N} \), which is an infinite subset of a discrete metric space \( \mathbb{N} \) where \( \rho(x, y) = 1 \) for \( x eq y \).
04

Checking Boundedness of Set A

To check if the set \( \mathbb{N} \) is bounded, we find that for any point in \( \mathbb{N} \) and any fixed distance greater than zero, the entire set will always be within that distance because no finite maximum exists for the entire set \( \mathbb{N} \). Thus in a loose sense of bounded, this allows it to be called "bounded".
05

Disproving Total Boundedness

Now, we determine if \( \mathbb{N} \) is totally bounded with some \( \varepsilon = 0.5 \). In a discrete space, the balls of radius \( \varepsilon < 1 \) around a point will only include the point itself because no other point within distance less than 1 exists. Therefore, no finite number of balls can cover \( \mathbb{N} \) because it’s infinite and each ball only covers one point.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Totally Bounded
Understanding *totally bounded* sets is key to grasping certain concepts in metric spaces. A set is considered totally bounded if, regardless of how tiny a radius you choose, you can cover the set with a finite number of open balls (or *globes*). Imagine trying to cover a shape with the least number of equally-sized circles; if you can manage it no matter the circle's size, your set is totally bounded.

For example, consider the set of rational numbers between 0 and 1. No matter how small the radius, you can always find a finite number of intervals to cover this set. It's like tucking a blanket around them regardless of how the temperature changes. But note, being totally bounded is a stricter condition than just being bounded. It means you can "bunch up" the set into these small covers completely and finitely.

Key points:
  • Requires finite coverage with small radius.
  • Stricter than mere boundedness.
Discrete Metric Space
A *discrete metric space* is a special and simpler type of metric space where distances between distinct points are clear cut. In this space, the distance between any two distinct points is defined to be 1, while the distance from a point to itself is actually 0.

This setup makes the concept of distance straightforward. In such a setting, every subset is both closed and open as each point essentially stands alone with respect to its neighbors. This distinct sharpness of distance helps make many problems about distance calculations easier.

It's worth noting that in such spaces, visualizing concepts like total boundedness or boundedness might differ compared to continuous spaces:
  • Distinct points: distance 1.
  • Same points: distance 0.
  • Simplifies many distance complexities.
Bounded Sets
A set is labeled as *bounded* in a metric space if you can enclose it within some finite distance from a specific point. Imagine having a boundary or a fence around your set; if you can find such a fence with a fixed width, then your set is bounded.

This concept doesn't consider the number of elements or size of the set, only if you can stop it from "stretching out" indefinitely. Even an infinite set can be called bounded if it fits snugly inside such a fence. But remember, being bounded doesn't necessarily make a set totally bounded, as seen in some infinite sets in discrete spaces.

Key attributes:
  • Contained within a fixed distance.
  • Doesn’t extend beyond a certain point.
  • Can include infinite sets.
Infinite Sets
When dealing with *infinite sets*, the rules and intuition that apply to finite sets often need adjustment. An infinite set has no end; it keeps going, continuing without closure. A prime example is the set of natural numbers, which goes on indefinitely.

In metric spaces, infinite sets can still hold peculiar properties. For instance, in a discrete metric space, an infinite set can be bounded but not totally bounded. This seems paradoxical but highlights the importance of understanding different types of *boundedness*.

Attributes of infinite sets:
  • No finite end.
  • Common example: natural numbers.
  • Can exemplify different boundedness properties.

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Most popular questions from this chapter

Prove that $$ |\rho(y, z)-\rho(x, z)| \leq \rho(x, y) $$ in any metric space \((S, \rho)\).

Show that \(E^{2}\) becomes a metric space if distances \(\rho(\bar{x}, \bar{y})\) are defined by (a) \(\rho(\bar{x}, \bar{y})=\left|x_{1}-y_{1}\right|+\left|x_{2}-y_{2}\right|\) or (b) \(\rho(\bar{x}, \bar{y})=\max \left\\{\left|x_{1}-y_{1}\right|,\left|x_{2}-y_{2}\right|\right\\}\) where \(\bar{x}=\left(x_{1}, x_{2}\right)\) and \(\bar{y}=\left(y_{1}, y_{2}\right) .\) In each case, describe \(G_{\overline{0}}(1)\) and \(S_{\overline{0}}(1) .\) Do the same for the subspace of points with nonnegative coordinates.

Take for granted the lemma that $$ a^{1 / p} b^{1 / q} \leq \frac{a}{p}+\frac{b}{q} $$ if \(a, b, p, q \in E^{1}\) with \(a, b \geq 0\) and \(p, q>0,\) and $$ \frac{1}{p}+\frac{1}{q}=1 $$ (A proof will be suggested in Chapter \(5, \S 6,\) Problem 11.) Use it to prove Hölder's inequality, namely, if \(p>1\) and \(\frac{1}{p}+\frac{1}{q}=1,\) then $$ \sum_{k=1}^{n}\left|x_{k} y_{k}\right| \leq\left(\sum_{k=1}^{n}\left|x_{k}\right|^{p}\right)^{\frac{1}{p}}\left(\sum_{k=1}^{n}\left|y_{k}\right|^{q}\right)^{\frac{1}{q}} \text { for any } x_{k}, y_{k} \in C . $$

Prove (for \(E^{2}\) and \(E^{3}\) ) that $$ \bar{x} \cdot \bar{y}=|\bar{x}||\bar{y}| \cos \alpha $$ where \(\alpha\) is the angle between the vectors \(\overrightarrow{0 x}\) and \(\overrightarrow{0 y}\); we denote \(\alpha\) by \(\langle\bar{x}, \bar{y}\rangle .\)

Prove that \(\bar{A}\) is contained in any closed superset of \(A\) and is the intersection of all such supersets.

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