Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Show that each arc is a continuous image of [0,1] . [Hint: First, show that any \([a, b] \subseteq E^{1}\) is such an image. Then use a suitable composite mapping.]

Short Answer

Expert verified
An arc is a continuous image of \([0,1]\) via a suitable continuous function.

Step by step solution

01

Understanding the problem

The problem asks us to show that any given arc can be represented as a continuous image of the interval ([0, 1]). This means that for any arc, we should identify a continuous function \(f: [0,1] \to \mathbb{R}^{n}\) such that \(f([0,1])\) represents the arc itself.
02

Prove basic intervals are continuous images

Consider a closed interval \([a, b]\subseteq \mathbb{R}^{1}\). Define a function \(f: [0,1] \to [a,b]\) by the linear mapping\[f(t) = a + (b-a)t\]This function maps 0 to \(a\) and 1 to \(b\), covering every point in between as \(t\) varies from 0 to 1. Since \(f(t)\) is linear, it is continuous, showing that \([a, b]\) is a continuous image of \([0,1]\).
03

Generalize to an arc in higher dimensions

An arc in \(\mathbb{R}^{n}\) can be described by a continuous mapping \(g: [0,1] \to \mathbb{R}^{n}\). Suppose you have a parameterization of an arc, \(\gamma : [0,1] \to \mathbb{R}^n\), which represents the arc as a continuous function over \([0,1]\). Therefore, if any line segment in \(\mathbb{R}^{1}\) is a continuous image of \([0,1]\), we can extend this idea to represent a continuous path (or arc) in higher-dimensional space.
04

Use composition for broader mapping

To satisfy the requirement of demonstrating that an arc is a continuous image of \([0, 1]\), consider using a composite function. Let\[f: [0, 1] \to \text{Image of arc}\]be defined as a composition of the re-parametrizing function (say, mapping \([0, 1]\) to some \([a, b]\)) and the arc's parameterization from \([a, b]\) onto the arc in \(\mathbb{R}^{n}\). The composition \(\gamma \circ f: [0,1] \to \mathbb{R}^n\) will be continuous since it is the composition of continuous functions.

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 Vaia!

Key Concepts

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

Arc Representation
An arc in mathematics generally refers to a curve that connects two points smoothly. To demonstrate that an arc can be represented as the continuous image of the interval \([0, 1]\), we start with the concept of mapping a simple interval in one-dimensional space to the arc itself.

The task involves constructing a function that can take every value on \([0, 1]\) and map it to points on the arc. This involves defining a continuous function \(f: [0,1] \to \mathbb{R}^{n}\) such that the image of the function covers the entire arc.

To achieve this, we begin by representing basic line segments \([a, b]\) in \(\mathbb{R}^{1}\) using a linear function. Then, extend this representation through composite functions to handle more complex and higher-dimensional arcs. The base idea is that any shape that corresponds to an arc can be generated from the interval \([0, 1]\) through a series of transformations.
Real Analysis
Real Analysis often deals with functions, limits, continuity, and mappings. In the context of arc representation, we leverage real analysis to formalize and prove that an interval can map onto an arc continuously.

The particular interest here is in understanding how continuous functions behave. A function \(f: [0,1] \to [a,b]\) is continuous if small changes in its input (values from \([0, 1]\)) result in small changes in the output (values in the arc). Thus, real analysis provides the theoretical groundwork for treating the continuity of such mappings.

Properties of continuity are crucial because they ensure there are no jumps or breaks as we move from 0 to 1 on the interval. A continuous image, in this case, means that as we traverse the interval \([0,1]\), we trace out the arc without any disconnections. This concept is essential in proving the existence of a continuous mapping that can take one-dimensional intervals and extend them into continuous paths.
Higher Dimensional Mapping
Mapping in higher dimensions builds upon the basic idea of creating continuous images of arcs. Here, we deal with extending one-dimensional concepts like line segments into two or more dimensions to form curves, surfaces, and more complex geometries.

When dealing with mappings in \(\mathbb{R}^{n}\), arcs are represented as images of continuous functions from the interval \([0, 1]\). These functions are parameterized such that every value in \([0, 1]\) corresponds to a unique point on the n-dimensional arc.

To construct such mappings, we utilize compositions of simple mappings. By combining a linear mapping from \([0, 1]\) to \([a, b]\) and another mapping that depicts the arc in higher-dimensional space, we achieve continuous images in higher dimensions. This idea is central to many topics in geometry and analysis, such as the study of curves, spaces, and smooth structures.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Show that every polynomial of degree one on \(E^{n}\left({ }^{*}\right.\) or \(\left.C^{n}\right)\) is uniformly continuous.

Prove that the convergence or divergence (pointwise or uniformly) of a sequence \(\left\\{f_{m}\right\\}\), or a series \(\sum f_{m}\), of functions is not affected by deleting or adding a finite number of terms. Prove also that \(\lim _{m \rightarrow \infty} f_{m}\) (if any) remains the same, but \(\sum_{m=1}^{\infty} f_{m}\) is altered by the difference between the added and deleted terms.

Let $$ f(x)=\sum_{k=0}^{n} a_{k} x^{k} \text { and } g(x)=\sum_{k=0}^{m} b_{k} x^{k}\left(a_{n} \neq 0, b_{m} \neq 0\right) $$ Find \(\lim _{x \rightarrow \infty} \frac{f(x)}{g(x)}\) if (i) \(n>m ;\) (ii) \(n

Prove that if two functions \(f, g\) with values in a normed vèctor space are uniformly continuous on a set \(B,\) so also are \(f \pm g\) and \(a f\) for a fixed scalar \(a\). For real functions, prove this also for \(f \vee g\) and \(f \wedge g\) defined by $$(f \vee g)(x)=\max (f(x), g(x))$$ and $$ (f \wedge g)(x)=\min (f(x), g(x)) $$

In the following cases, find \(\lim f(x)\) in two ways: (i) use definitions only; (ii) use suitable theorems and justify each step accordingly. (a) \(\lim _{x \rightarrow \infty} \frac{1}{x}(=0)\). (b) \(\lim _{x \rightarrow \infty} \frac{x(x-1)}{1-3 x^{2}}\). (c) \(\lim _{x \rightarrow 2^{+}} \frac{x^{2}-2 x+1}{x^{2}-3 x+2}\). (d) \(\lim _{x \rightarrow 2^{-}} \frac{x^{2}-2 x+1}{x^{2}-3 x+2}\). (e) \(\lim _{x \rightarrow 2} \frac{x^{2}-2 x+1}{x^{2}-3 x+2}(=\infty)\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free