/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 Let \(I:=[a, b]\) and let \(f: I... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 10

Let \(I:=[a, b]\) and let \(f: I \rightarrow \mathbb{R}\) and \(g: I \rightarrow \mathbb{R}\) be continuous functions on \(I\). Show that the set \(\\{x \in I: f(x)=g(x)\\}\) is closed in \(\mathbb{R}\).

Short Answer

Expert verified
The set \(S = \{x \in I: f(x)=g(x)\}\) is closed in \(\mathbb{R}\) because the difference between two continuous functions \(f\) and \(g\) is continuous and the preimage of a closed set under a continuous function is also closed.

Step by step solution

01

Show that \(f - g\) is continuous

By the properties of continuous functions, the difference between two continuous functions, \(f\) and \(g\), is also a continuous function. We can define a function \(h(x) = f(x) - g(x)\). Since \(f\) and \(g\) are given to be continuous on \(I\), so is their difference, \(h\). Therefore, \(h: I \rightarrow \mathbb{R}\) is continuous.
02

Define the set we want to prove is closed

Now define the set \(S = \{x \in I: f(x)=g(x)\}\). This can also be written as \(S = \{x \in I: h(x)=0\}\). We want to show that this set \(S\) is closed in \(\mathbb{R}\).
03

Use the property of continuous functions

By the properties of continuous functions, the preimage of a closed set under a continuous function is closed. In this case, \(h^{-1}(\{0\}) = S\) is the preimage of the closed set \(\{0\}\) under the continuous function \(h\). Therefore, \(S\) is a closed set in \(\mathbb{R}\).

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 91影视!

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

Let \(I:=[1, \infty)\) and let \(f(x):=\sqrt{x-1}\) for \(x \in I\). For each \(\varepsilon\) -neighborhood \(G=(-\varepsilon .+\varepsilon)\) of 0\. exhibit an open set \(H\) such that \(H \cap I=f^{-1}(G)\).

Show that a set \(G \subseteq \mathbb{R}\) is open if and only if it does not contain any of its boundary points.

Let \(K \neq \emptyset\) be a compact set in \(\mathbb{R}\). Show that inf \(K\) and sup \(K\) exist and belong to \(K\).

Prove that in any metric space, an \(\varepsilon\) -neighborhood of a point is an open set.

Show that if \(f: \mathbb{R} \rightarrow \mathbb{R}\) is continuous, then the set \(\\{x \in \mathbb{R}: f(x)<\alpha\\}\) is open in \(\mathbb{R}\) for cach \(\alpha \in \mathbb{R}\).
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation
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.