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Chapter 12: Problem 12

Man w盲hle in einer cattesischen Ebene cinen festen Punkt \(x_{0}\) (und nenne ihn 鈥濸aris"). Definiere den Abstand \(\mathrm{d}(\boldsymbol{x}, \boldsymbol{y})\) zwischen den Punkten \(\boldsymbol{x}, \boldsymbol{y}\) folgenderma脽en: \(\mathrm{d}(\boldsymbol{x}, \mathbf{y})\) sei der euklidische Abstand zwischen \(\boldsymbol{x}\) und \(\mathbf{y}\), falls diese beiden Punkte auf ciner Geraden durch \(x_{0}\) liegen; andernfalls sei \(\mathrm{d}(x, y)\) die Summe der euklidischen Abst盲nde 2wischen \(x, x_{0}\) und \(y, x_{0}\). Zeige, daB hierdurch eine Metrik auf \(R^{2}\) erkl?rt wird (,,Metrik des franz枚sischen Eisenbahnsystems " \(^{\text {t'}) . ~}\)

Short Answer

Expert verified
The metric satisfies positivity, symmetry, and the triangle inequality.

Step by step solution

01

Define the Metric Conditions

For any metric, three conditions must be satisfied: positivity, symmetry, and the triangle inequality. Let's define these conditions:1. Positivity: \(d(x, y) \geq 0\) and \(d(x, y) = 0\) if and only if \(x = y\).2. Symmetry: \(d(x, y) = d(y, x)\) for all points \(x\) and \(y\).3. Triangle Inequality: For any points \(x, y, z\), \(d(x, y) \leq d(x, z) + d(z, y)\).
02

Verify Positivity

Consider two cases based on the given distance definition:1. If \(x\) and \(y\) are on the same line through \(x_0\), the distance is the Euclidean distance \(d(x, y) = ||x - y||\), which is always non-negative and zero if and only if \(x = y\).2. If \(x\) and \(y\) are not on the same line through \(x_0\), then \(d(x, y) = ||x - x_0|| + ||y - x_0||\), which is non-negative and zero if and only if both \(x = x_0\) and \(y = x_0\), which implies \(x = y\). Thus, positivity is satisfied.
03

Verify Symmetry

The distance is symmetric by construction:1. If \(x\) and \(y\) are on the same line, \(d(x, y) = ||x - y|| = ||y - x|| = d(y, x)\).2. If \(x\) and \(y\) are not on the same line, \(d(x, y) = ||x - x_0|| + ||y - x_0|| = ||y - x_0|| + ||x - x_0|| = d(y, x)\).Hence, symmetry is satisfied.
04

Verify Triangle Inequality

Assess both cases against the triangle inequality:1. If \(x, y, z\) are collinear, the Euclidean triangle inequality applies directly: \(||x - y|| \leq ||x - z|| + ||z - y||\).2. If any two of the three points are not collinear with \(x_0\), say \(x\) and \(y\), then: \[d(x, y) = ||x - x_0|| + ||y - x_0|| \leq (||x - x_0|| + ||z - x_0||) + (||z - x_0|| + ||y - x_0||) = d(x, z) + d(z, y).\] Thus, the triangle inequality holds for both cases.

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Key Concepts

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

Euclidean Distance
Euclidean distance is a fundamental concept used to determine the straight-line distance between two points in a Euclidean space. In simple terms, it gives us the "as-the-crow-flies" distance. The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) in a two-dimensional space can be calculated using the formula:\[ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
  • This distance is always non-negative since it is derived from the square of differences, which are non-negative.
  • It equals zero if and only if the two points are identical.
Notably, in the given exercise, when two points \(x\) and \(y\) lie on a line through the fixed point \(x_0\), their distance is essentially the Euclidean distance. This intrinsic property of Euclidean space makes it a crucial part of metric spaces like the one described.
Triangle Inequality
The triangle inequality is an essential property of metrics that establishes a fundamental relationship in a metric space. For any three points \(x, y, z\) in such a space, the inequality states:\[d(x, y) \leq d(x, z) + d(z, y)\]This means that the direct distance between two points is always less than or equal to the sum of the distances when taking an intermediate detour.
  • In our exercise, if points \(x, y, z\) are all on a line through \(x_0\), the Euclidean version of this inequality holds using simple geometry.
  • For points not on the same line through \(x_0\), the exercise defines the distance as the sum of Euclidean distances 鈥� and still respects the triangle inequality.
This relation ensures that the metric function respects the structure of the space, providing consistency and predictability.
Metric Definition
The concept of a metric comes from the idea of measuring distance in a mathematical space. A metric must satisfy the following conditions to qualify as a true distance measurement:
  • Positivity: \(d(x, y) \geq 0\) and \(d(x, y) = 0\) if and only if \(x = y\). This simply means distances cannot be negative, and identical points have zero distance.
  • Symmetry: \(d(x, y) = d(y, x)\). Just like real-world distances, swapping the start and end points of a journey shouldn't change its length.
  • Triangle Inequality: Ensures the shortest path between two points is a direct line, maintaining consistency with our intuitive understanding of distance.
In the exercise related to the metric on \(\mathbb{R}^2\), these properties undergo verification:- Positivity is upheld whether points are collinear with \(x_0\) or not.- Symmetry naturally occurs thanks to the distance formula used.- The flexibility of distance definitions in the metric ensures that the triangle inequality holds true, securing its status as a metric space. Understanding these conditions helps us recognize and validate when a function truly measures distance.

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Most popular questions from this chapter

Sind die \(p_{k}\) Gewichte, also alle \(>0\), und ist \(a_{1}

F眉r nichtnegative \(a_{k}\) ist \(\min \left(a_{1}, \ldots, a_{n}\right) \leqslant\left(a_{1} \cdots a_{n}\right)^{1 / n} \leqslant \max \left(a_{1}, \ldots, a_{n}\right)\)

Ungleichung des quadratischen Mittels: F眉r nichtnegative \(a_{1}, \ldots, a_{n}\) ist \(\min \left(a_{1}, \ldots, a_{n}\right) \leqslant\left(\frac{a_{1}^{2}+\cdots+a_{n}^{2}}{n}\right)^{1 / 2} \leqslant \max \left(a_{1}, \ldots, a_{n}\right)\)

\(\frac{1}{\sqrt{n}} \sum_{k=1}^{n}\left|a_{k}\right| \leqslant\left(\sum_{k=1}^{n} a_{k}^{2}\right)^{1 / 2} \leq \sqrt{n} \max \left(\left|a_{1}\right| \ldots,\left|a_{n}\right|\right)\)

F? \(x>0\) ist \(x+\frac{1}{x} \geqslant 2\)
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