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

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 \(X\).

Short Answer

Expert verified
The metric \( \rho \) satisfies all the properties of a metric and defines the product topology on \( X \).

Step by step solution

01

Understanding the Metric Function

The metric function \( \rho(x, y) \) is defined as \( \rho(x, y) = \sum_{n=1}^{\infty} 2^{-n} \rho_{n}(x_{n}, y_{n}) \). The coefficients \( 2^{-n} \) ensure that the series converges, as each term \( \rho_{n}(x_{n}, y_{n}) \) is multiplied by a decreasing geometric factor.
02

Verifying Non-Negativity

For any elements \( x, y \) in the product space \( X \), the terms \( \rho_{n}(x_{n}, y_{n}) \) are metrics and thus non-negative. As a result, each term in the sum \( 2^{-n}\rho_{n}(x_{n}, y_{n}) \) is non-negative, making \( \rho(x, y) \geq 0 \).
03

Identity of Indiscernibles

The identity of indiscernibles property requires that \( \rho(x, y) = 0 \) if and only if \( x = y \). Since each \( \rho_{n} \) is a metric, \( \rho_{n}(x_{n}, y_{n}) = 0 \) if and only if \( x_{n} = y_{n} \) for all \( n \). Therefore, \( \rho(x, y) = 0 \) implies \( x = y \). Conversely, if \( x = y \), then \( \rho(x, y) = 0 \).
04

Symmetry of the Metric

The metric function \( \rho \) is symmetric if \( \rho(x, y) = \rho(y, x) \). Since each \( \rho_{n}(x_{n}, y_{n}) \) is symmetric, \( \rho_{n}(x_{n}, y_{n}) = \rho_{n}(y_{n}, x_{n}) \). Thus, \( \rho(x, y) = \sum_{n=1}^{\infty} 2^{-n}\rho_{n}(x_{n}, y_{n}) \) is equal to \( \sum_{n=1}^{\infty} 2^{-n} \rho_{n}(y_{n}, x_{n}) = \rho(y, x) \).
05

Proving the Triangle Inequality

For the triangle inequality, we show \( \rho(x, z) \leq \rho(x, y) + \rho(y, z) \) for any \( x, y, z \in X \). Each \( \rho_{n} \) satisfies the triangle inequality, i.e., \( \rho_{n}(x_{n}, z_{n}) \leq \rho_{n}(x_{n}, y_{n}) + \rho_{n}(y_{n}, z_{n}) \). Then: \[ \rho(x, z) = \sum_{n=1}^{\infty} 2^{-n}\rho_{n}(x_{n}, z_{n}) \leq \sum_{n=1}^{\infty} 2^{-n}(\rho_{n}(x_{n}, y_{n}) + \rho_{n}(y_{n}, z_{n})) = \rho(x, y) + \rho(y, z) \] As such, \( \rho \) satisfies the triangle inequality.
06

Conclusion on the Metric

Since the function \( \rho \) satisfies non-negativity, identity of indiscernibles, symmetry, and the triangle inequality, it is confirmed to be a metric. It defines the product topology on \( X \) because it matches the standard construction (sum of scaled component-wise distances) for product spaces.

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.

Product Topology
When discussing the product topology, we often speak about combining multiple topological spaces into a single cohesive structure. Imagine each space, denoted here as \(X_n\), has its own topology, which can be thought of as the "shape" or structure of that space. The product topology is created on a space that is the Cartesian product of these individual spaces. This means our new space is made up of all possible ordered tuples, where each element comes from one of the spaces \(X_n\).

The product topology ensures each component space influences the structure of our new, larger space. This is very much like how multiple layers of paint combine to give a final appearance. To rigorously construct the product topology, we use what is called a _basis_. The basis consists of all possible products of open sets from each individual component space. These sets look like rectangles in multi-dimensional spaces.

In the original exercise, we defined a metric \( \rho \) over this product space \( X = \prod_{1}^{\infty} X_{n} \). This metric is carefully constructed to promote convergence by weighting the influence of each space \( X_n \) using factors \( 2^{-n} \). The metric respects the structure of each component space while defining a global topology.
Triangle Inequality
The triangle inequality is an essential property of any metric space. This property states for any points \( x, y, z \) in a metric space, the distance from \( x \) to \( z \) should be no greater than the sum of the distances from \( x \) to \( y \) and \( y \) to \( z \). Formally, this is written as \( \rho(x, z) \leq \rho(x, y) + \rho(y, z) \).

In the context of our problem, each component space \( (X_n, \rho_n) \) satisfies the triangle inequality: \( \rho_n(x_n, z_n) \leq \rho_n(x_n, y_n) + \rho_n(y_n, z_n) \) for any points within that space. The overall metric \( \rho \) is designed to preserve this property across the entire product space.

By using the sum of scaled metrics \( \sum_{1}^{\infty} 2^{-n} \rho_{n}(x_{n}, z_{n}) \), any distance calculation on the product space respects the triangle inequality. Each component's metric reinforces the next, ensuring the full distance calculations adhere to this universal property. This ensures that our combined space behaves as a single, unified metric space.
Convergence of Series
Understanding the convergence of series is key when working with metrics in infinite-dimensional spaces like our product topology. In simple terms, a series converges if the sum of its infinite terms approaches a finite limit. Imagine adding smaller and smaller slices of pie - eventually, the size becomes negligible, and we approach a complete pie.

In our example, for the metric \( \rho(x, y) \) to be useful, the sum \( \sum_{n=1}^{\infty} 2^{-n} \rho_{n}(x_{n}, y_{n}) \) must converge. The geometric factor \( 2^{-n} \) plays a crucial role here. As \( n \) increases, the factor decreases exponentially, ensuring that terms corresponding to larger indices contribute less to the total sum. This is similar to the classic geometric series \( \sum_{n=1}^{\infty} 2^{-n} \) which converges to 1.

The concept of convergence guarantees our metric is well-defined and not infinite, allowing us to properly assess distances in our infinite-dimensional product space. The way these diminishing factors are structured ensures that the series keeps the metric bounded and practically useful.

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

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 \(X\) be a topological space equipped with an equivalence relation, \(\tilde{X}\) the set of equivalence classes, \(\pi: X \rightarrow \widetilde{X}\) the map taking each \(x \in X\) to its equivalence class, and \(\mathcal{T}=\left\\{U \subset \widetilde{X}: \pi^{-1}(U)\right.\) is open in \(\left.X\right\\} .\) a. \(\mathcal{J}\) is a topology on \(\widetilde{X}\). (It is called the quotient topology.) b. If \(Y\) is a topological space, \(f: \widetilde{X} \rightarrow Y\) is continuous iff \(f \circ \pi\) is continuous. c. \(\widetilde{X}\) is \(T_{1}\) iff every equivalence class is closed.

Let \(X\) be a topological space, \(Y\) a Hausdorff space, and \(f, g\) continuous maps from \(X\) to \(Y\). a. \(\\{x: f(x)=g(x)\\}\) is closed. b. If \(f=g\) on a dense subset of \(X\), then \(f=g\) on all of \(X\).

If \(X\) is a topological space, \(A \subset X\) is closed, and \(g \in C(A)\) satisfies \(g=0\) on \(\partial A\), then the extension of \(g\) to \(X\) defined by \(g(x)=0\) for \(x \in A^{c}\) is continuous.

If \(X\) is normal and second countable, there is a countable family \(\mathcal{F} \subset C(X, I)\) that separates points and closed sets. (Let \(\mathcal{B}\) be a countable base for the topology. Consider the set of pairs \((U, V) \in \mathcal{B} \times \mathcal{B}\) such that \(\bar{U} \subset V\), and use Urysohn's lemma.)

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