Logout Open In App
Open In App
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

Chapter 4: Problem 21

If \(X\) is an infinite set with the cofinite topology, then every \(f \in C(X)\) is constant.

Short Answer

Expert verified
Every function in \(C(X)\) is constant due to the cofinite topology's properties.

Step by step solution

01

Understand Cofinite Topology

The cofinite topology on a set \(X\) is a topology where the open sets are either \(X\) itself or complements of finite subsets of \(X\). This means all open sets except \(X\) are the entire space minus a finite number of points.
02

Define Continuous Functions on Topological Spaces

A function \(f: X \to Y\) between topological spaces is continuous if for every open set \(V\) in \(Y\), the preimage \(f^{-1}(V)\) is open in \(X\). In the cofinite topology, this implies that the preimage of any open set in \(Y\) must be cofinite or the entire space \(X\).
03

Show Preimage Condition Forces Constancy

Consider a continuous function \(f: X \to \mathbb{R}\). Assume \(f\) is not constant, and let \(a\) and \(b\) be distinct values in \(f(X)\). Choose open sets \(U, V \subset \mathbb{R}\) such that \(a \in U\) and \(b \in V\) with no overlap. \(f^{-1}(U)\) and \(f^{-1}(V)\) are cofinite, but their union is the whole space, which is not possible since \(f^{-1}(U) \cap f^{-1}(V) = \emptyset\). This contradiction implies \(f\) must be constant.
04

Conclude Proof

Since attempting to define a non-constant function leads to a contradiction under the cofinite topology, it follows that every continuous function \(f\) on \(X\) must map to a single point, hence every \(f \in C(X)\) is constant.

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.

Start your free trial

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.

Continuous Functions
Continuous functions play a crucial role in topology as they preserve the structure of topological spaces. To understand a continuous function, consider a function \( f: X \to Y \) between two topological spaces \(X\) and \(Y\). This function is deemed continuous if the preimage (the inverse image) of every open set in \(Y\) is also an open set in \(X\).
For example, picturing you have a rubber sheet mapped onto another surface. If you stretch or compress it without tearing, you have a continuous map. It's like tracing paths with no sharp breaks or jumps. The beauty of continuous functions is how they maintain the 'openness' of sets, ensuring a smooth transition from one part of \(X\) to another part of \(Y\).
In the context of cofinite topology, this requirement for continuity means very specific behavior, often leading to constancy. This occurs because the structure of open sets in the cofinite topology restricts the possible ways the function can map the elements without losing continuity.
Topological Spaces
Topological spaces are foundational to modern mathematics and can be thought of as a set \(X\) equipped with a topology \( \tau \). This topology is simply a collection of open sets that satisfy certain axioms, enabling complex geometric and analytic theories.
A good way to visualize a topological space is to think about different ways of interpreting the structure of a set of points—like connecting them in various ways to form different shapes. These shapes must follow rules such as the union or intersection of open sets being an open set, and the whole set, as well as the empty set, being open.
In the scenario of cofinite topology, it's about the vast majority of points being inside our open sets, with allowances made for only finitely many points to be excluded. Such a topology is particularly used to study properties of infinite sets and their behavior in terms of continuity, convergence, and more. It is quite abstract but elegantly ties in numerous concepts from other areas of mathematics.
Infinite Set
Infinite sets are collections of elements that don't have a finite size—they go on and on. A simple example is the set of all natural numbers \( \{1, 2, 3, \ldots\} \). Infinite sets can be categorized in various ways, such as countable or uncountable, depending on their size relative to the set of natural numbers.
In topology, when dealing with infinite sets, the notion of size and finiteness become crucial. The cofinite topology is one special case where the finiteness of the complement of open sets defines the topology. It limits open sets to being either the whole infinite set or the removal of just a finite number of elements from it.
This characteristic leads to interesting consequences in continuity and mapping, like in our original problem, where any continuous function from an infinite set in cofinite topology must be constant. Such ideas highlight the interconnectedness of these fundamental concepts in mathematics, showcasing how topology provides lenses to view infinite sets under unique properties and rules.

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 \(X\) be a compact Hausdorff space. An ideal in \(C(X, \mathbb{R})\) is a subalgebra \(\mathcal{J}\) of \(C(X, \mathbb{R})\) such that if \(f \in J\) and \(g \in C(X, \mathbb{R})\) then \(f g \in J\). a. If \(J\) is an ideal in \(C(X, \mathbb{R})\), let \(h(\mathcal{J})=\\{x \in X: f(x)=0\) for all \(f \in \mathcal{J}\\} .\) Then \(h(\mathcal{J})\) is a closed subset of \(X\), called the hull of J. b. If \(E \subset X\), let \(k(E)=\\{f \in C(X, \mathbb{R}): f(x)=0\) for all \(x \in E\\}\). Then \(k(E)\) is a closed ideal in \(C(X, \mathbb{R})\), called the kernel of \(E\). c. If \(E \subset X\), then \(h(k(E))=\bar{E}\). d. If \(J\) is an ideal in \(C(X, R)\), then \(k(h(\mathcal{J}))=J\). ( Hint: \(k(h(J))\) may be identified with a subalgebra of \(C_{0}(U, \mathbb{R})\) where \(\left.U=X \backslash h(J) .\right)\) e. The closed subsets of \(X\) are in one-to-one correspondence with the closed ideals of \(C(X, \mathbb{R})\).

Let \(\left\\{\left(X_{n}, \rho_{n}\right)\right\\}_{1}^{\infty}\) be a countable family of metric spaces whose metrics take values in \([0,1]\). (The latter restriction can always be satisfied; see Exercise \(56 \mathrm{~b}\).) Let \(X=\prod_{1}^{\infty} X_{n .}\) If \(x, y \in X\), say \(x=\left(x_{1}, x_{2}, \ldots\right)\) and \(y=\left(y_{1}, y_{2}, \ldots\right)\), define \(\rho(x, y)=\sum_{1}^{\infty} 2^{-n} \rho_{n}\left(x_{3}, y_{n}\right) .\) Then \(\rho\) is a metric that defines the product topology on \(X\).

If \(\operatorname{card}(X) \geq 2\), there is a topology on \(X\) that is \(T_{0}\) but not \(T_{1}\).

If \(A\) is a directed set, a subset \(B\) of \(A\) is called cofinal in \(A\) if for each \(\alpha \in A\) there exists \(\beta \in B\) such that \(\beta \gtrsim \alpha\). a. If \(B\) is cofinal in \(A\) and \(\left\langle x_{\alpha}\right\rangle_{\alpha \in A}\) is a net, the inclusion map \(B \rightarrow A\) makes \(\left\langle x_{\beta}\right\rangle_{\beta \in B}\) a subnet of \(\left\langle x_{\alpha}\right\rangle_{\alpha \in A}\). b. If \(\left\langle x_{\alpha}\right\rangle_{\alpha} \in A\) is a net in a topological space, then \(\left\langle x_{\alpha}\right\rangle\) converges to \(x\) iff for every cofinal \(B \subset A\) there is a cofinal \(C \subset B\) such that \(\left\langle x_{\gamma}\right\rangle_{\gamma \in C}\) converges to \(x\).

Let \(Q\) have the relative topology induced from \(R\). a. \(\mathbb{Q}\) is not locally compact. b. \(\mathbb{Q}\) is \(\sigma\)-compact (it is a countable union of singleton sets), but uniform convergence on singletons (i.e., pointwise convergence) does not imply uniform convergence on compact subsets of \(Q\).
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

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.

Sign-up for free