Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 13

Let \(X\) and \(Y\) be topological spaccs and \(f . X \times Y \rightarrow X\) a continuous map. For each fixed \(a \in X\) show that the map \(f_{n}: Y \rightarrow X\) defined by \(f_{\theta}(v)=f(a, v)\) is contunous.

Short Answer

Expert verified
Yes, with each fixed \(a \in X\), the map \(f_{\theta}: Y \rightarrow X\) defined by \(f_{\theta}(v)=f(a, v)\) is continuous.

Step by step solution

01

Define the function

The first step is to define the function \(f_{\theta}: Y \rightarrow X\), which is given by \(f_{\theta}(v) = f(a, v)\) where \(a\) is a fixed point in \(X\). By definition, for each \(a \in X\), we have a function \(f_{\theta}: Y \rightarrow X\).
02

Check the pre-image of an open subset

The next step is to check for any open subset \(U\) of \(X\), the pre-image under \(f_{\theta}\), i.e., \(f_{\theta}^{-1}(U)\) is open in \(Y\). Here \(f_{\theta}^{-1}(U) = \{v \in Y : f_{\theta}(v) \in U \}\).
03

Verify whether the pre-image is open

Observe that \(f_{\theta}^{-1}(U) = \{v \in Y : f_{\theta}(v) \in U \} = \{v \in Y : f(a, v) \in U \}\). This set is actually equal to \(f^{-1}(U \times Y)\), where \(f: X \times Y \rightarrow X\) is the original continuous map. Since \(f\) is continuous and \(U \times Y\) is open in \(X \times Y\), \(f^{-1}(U \times Y)\) is open in \(X \times Y\). Hence, \(f_{\theta}^{-1}(U)\) which is a subset of \(f^{-1}(U \times Y)\) is open in \(Y\). Hence \(f_{\theta}: Y \rightarrow X\) is continuous.

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!

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

A topological space \(X\) is called normal if for evcry pair of disjo?nt closcd suhsets \(A\) and \(B\) there exist disjoint open sets \(U\) and \(V\) sich that \(A \subset U\) and \(B \subset V\) Show that every metric space is nomal

Show that if \(x_{n}\) is a sequence in anormed vector space \(V\) such that every subsequence has a subsequence comergent to \(x\), then \(x_{n} \rightarrow x\).

Let \(V\) be a Banach space and \(W\) te a vector subspace of \(V .\) Define its closture \(\bar{W}\) to be the union of \(W\) and all hmuts of Cauchy sequences of elements of \(W\). Show that \(\bar{W}\) is a closed vector subspace of \(V\) in the sense that the limit points of all Cauchy scquences in \(\bar{W}\) lie in \(\bar{W}\) (note that the Cauchy sequences may include the newly added limit points of \(W\) ).

Show that cvery space \(F(S)\) is complete with respect to the supremum norm of Example 10.26. Hence show that the vector space \(\ell_{\infty}\) of bounded infinite complex sequences is a Banach space with respect to the norm \(\|\mathrm{x}\|=\sup \left(x_{t}\right)\).

If \(G_{0}\) is the component of the identity of a locally connected topological group \(G\), the factor group \(G / G_{0}\) is called the group of components of \(G .\) Show that the group of components is a discrete topological group with respect to the topology induced by the natural projection map \(\pi: g \mapsto g G_{0}\)
See all solutions

Recommended explanations on Combined Science Textbooks

Synergy

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