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Chapter 3: Problem 10

Prove the following about lines and line segments. (i) Show that any line segment in \(E^{n}\) is a bounded set, but the entire line is not. (ii) Prove that the diameter of \(L(\bar{a}, \bar{b})\) and of \((\bar{a}, \bar{b})\) equals \(\rho(\bar{a}, \bar{b})\).

Short Answer

Expert verified
(i) Line segments are bounded; lines are not. (ii) Diameters are \(\rho(\bar{a}, \bar{b})\) for both open and closed segments.

Step by step solution

01

Understanding Line Segment as a Bounded Set

A line segment in a Euclidean space \(E^n\) is defined between two points \(\bar{a}\) and \(\bar{b}\). To prove that this is a bounded set, we recognize that all points on the line segment \(L(\bar{a}, \bar{b})\) must satisfy:\[ \bar{x} = (1-t)\bar{a} + t\bar{b} \] where \(0 \leq t \leq 1\). The distance between any point \(\bar{x}\) on the segment and either endpoint \(\bar{a}\) or \(\bar{b}\) is finite and does not exceed \(\rho(\bar{a}, \bar{b})\). Hence, the line segment is bounded.
02

Understanding an Entire Line as Unbounded

An entire line in \(E^n\) extends infinitely in both directions. For any two points \(\bar{a}\) and \(\bar{b}\) on a line, the line continues beyond these points indefinitely. Therefore, for every distance \(R\), you can find a point \(\bar{x}\) on the line where the distance \(\rho(\bar{x}, \bar{a}) > R \), proving that the entire line is not a bounded set.
03

Proving Diameter of a Closed Line Segment

The diameter of a set is the greatest distance between any two points in the set. For the closed line segment \(L(\bar{a}, \bar{b})\), the endpoints \(\bar{a}\) and \(\bar{b}\) are the furthest apart, making the diameter equal to \(\rho(\bar{a}, \bar{b})\).
04

Proving Diameter of an Open Line Segment

An open line segment \((\bar{a}, \bar{b})\) includes all points between \(\bar{a}\) and \(\bar{b}\) but not the endpoints themselves. However, any two points very close to \(\bar{a}\) and \(\bar{b}\) can have a distance approaching \(\rho(\bar{a}, \bar{b})\). Thus, the diameter of \((\bar{a}, \bar{b})\) also equals \(\rho(\bar{a}, \bar{b})\).

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Key Concepts

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

Bounded Sets
In Euclidean Geometry, a bounded set is a collection of points that have a finite distance in space. Imagine a boundary that keeps all points within a certain area. A simple way to visualize this is to think about a circle or a rectangle. Every dot within these shapes is "bounded" because it is contained by the edges of the shape.

When we apply this to the concept of a line segment, we see that the segment is made up of points lying between two fixed endpoints. For example, in the Euclidean space \(E^n\), a line segment \(L(\bar{a}, \bar{b})\) connects the points \(\bar{a}\) and \(\bar{b}\), and all the points on the segment can be expressed as:
  • \(\bar{x} = (1-t)\bar{a} + t\bar{b}\)
where \(0 \leq t \leq 1\). Each of these points is at a finite distance from either \(\bar{a}\) or \(\bar{b}\). Hence they are bounded.

In contrast, an entire line extends infinitely. There's no set boundary to stop it. Thus, it is unbounded, as you can wander endlessly in either direction without reaching an end.
Line Segments
A line segment in geometry is a part of a line that is bounded by two distinct end points. It is not infinite like a line, but instead starts at one point and stops at another. This is what makes it a fundamental concept in Euclidean space.

The distinguishing features of a line segment are:
  • It has two endpoints (like \(\bar{a}\) and \(\bar{b}\))
  • It includes all points that lie directly between these endpoints
  • It has a measurable length, known as its "distance" or "diameter"
This length is what separates a line segment from a full line, which goes on forever. Understanding line segments is essential for comprehending more complex geometric structures.
Diameter of a Set
The diameter of a set in Euclidean geometry represents the largest distance you can measure between any two points within a given set. If you were to imagine fitting the longest possible ruler within the set, the measurement it reads would be the diameter.

For a closed line segment \(L(\bar{a}, \bar{b})\), the diameter is simply the distance between its endpoints, \(\bar{a}\) and \(\bar{b}\). This is mathematically expressed as \(\rho(\bar{a}, \bar{b})\), indicating the measure directly from \(\bar{a}\) to \(\bar{b}\).

Now, even if the line segment is open, meaning it does not include the points \(\bar{a}\) and \(\bar{b}\) themselves but consists of all points between them, the diameter remains the same. Our points of measure can get infinitely close to the endpoints, thus the calculated diameter continues to approach \(\rho(\bar{a}, \bar{b})\). This is because no other pair of points in the segment could have a greater distance between them than the ones hugging \(\bar{a}\) and \(\bar{b}\).

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Most popular questions from this chapter

Define $$ e^{\theta i}=\cos \theta+i \sin \theta $$ Describe \(e^{\theta i}\) geometrically. Is \(\left|e^{\theta i}\right|=1 ?\)

Prove that if \(\left(X, \rho^{\prime}\right)\) and \(\left(Y, \rho^{\prime \prime}\right)\) are metric spaces, then a metric \(\rho\) for the set \(X \times Y\) is obtained by setting, for \(x_{1}, x_{2} \in X\) and \(y_{1}, y_{2} \in Y\), (i) \(\rho\left(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)\right)=\max \left\\{\rho^{\prime}\left(x_{1}, x_{2}\right), \rho^{\prime \prime}\left(y_{1}, y_{2}\right)\right\\}\); or (ii) \(\rho\left(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)\right)=\sqrt{\rho^{\prime}\left(x_{1}, x_{2}\right)^{2}+\rho^{\prime \prime}\left(y_{1}, y_{2}\right)^{2}}\)

Find the edge-lengths of \(A=(\bar{a}, \bar{b})\) in \(E^{4}\) if $$ \bar{a}=(1,-2,4,0) \text { and } \bar{b}=(2,0,5,3) $$ Is \(A\) a cube? Find some rational points in it. Find \(d A\) and \(v A\).

Prove the additivity of the volume of intervals, namely, if \(A\) is subdivided, in any manner, into \(m\) mutually disjoint subintervals \(A_{1}, A_{2}, \ldots, A_{m}\) in \(E^{n}\), then $$ v A=\sum_{i=1}^{m} v A_{i} $$ (This is true also if some \(A_{i}\) contain common faces). [Proof outline: For \(m=2\), use Problem 8 Then by induction, suppose additivity holds for any number of intervals smaller than a certain \(m\) \((m>1) .\) Now let $$ A=\bigcup_{i=1}^{m} A_{i} \quad\left(A_{i} \text { disjoint }\right) $$ One of the \(A_{i}\) (say, \(\left.A_{1}=[\bar{a}, \bar{p}]\right)\) must have some edge-length smaller than the corresponding edge-length of \(A\left(\right.\) say \(\left., \ell_{1}\right) .\) Now cut all of \(A\) into $$ P=[\bar{a}, \bar{d}] \text { and } Q=A-P \text { (Figure 4) } $$ by the plane \(x_{1}=c\left(c=p_{1}\right)\) so that \(A_{1} \subseteq P\) while \(A_{2} \subseteq Q\). For simplicity, assume that the plane cuts each \(A_{i}\) into two subintervals \(A_{i}^{\prime}\) and \(A_{i}^{\prime \prime}\). (One of them may be empty.) Then $$ P=\bigcup_{i=1}^{m} A_{i}^{\prime} \text { and } Q=\bigcup_{i=1}^{m} A_{i}^{\prime \prime} $$ Actually, however, \(P\) and \(Q\) are split into fewer than \(m\) (nonempty) intervals since \(A_{1}^{\prime \prime}=\emptyset=A_{2}^{\prime}\) by construction. Thus, by our inductive assumption, $$ v P=\sum_{i=1}^{m} v A_{i}^{\prime} \text { and } v Q=\sum_{i=1}^{m} v A_{i}^{\prime \prime} $$ where \(v A_{1}^{\prime \prime}=0=v A_{2}^{\prime},\) and \(v A_{i}=v A_{i}^{\prime}+v A_{i}^{\prime \prime}\) by Problem \(8 .\) Complete the inductive proof by showing that $$ \left.v A=v P+v Q=\sum_{i=1}^{m} v A_{i} .\right] $$

Compute (a) \(\frac{1+2 i}{3-i}\) (b) \((1+2 i)(3-i) ;\) and (c) \(\frac{x+1+i}{x+1-i}, x \in E^{1}\). Do it in two ways: (i) using definitions only and the notation \((x, y)\) for \(x+y i\); and (ii) using all laws valid in a field.
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