**Prove that for any infinite index set $ J $, the uniform topology on $ \mathbb{R}^J $ is strictly finer than the product topology on $ \mathbb{R}^J $.**This statement involves comparing two different topologies on the product space $ \mathbb{R}^J $, where $ J $ is an infinite index set:1. **Product Topology**: This topology is generated by the basis of sets that are products of open intervals in $\mathbb{R}$, with all but finitely many factors being the entire real line $\mathbb{R}$.2. **Uniform Topology**: This topology corresponds to the convergence of functions from $ J $ to $\mathbb{R}$ uniformly, that is, where the difference between function values is small across the entire index set $ J $.A topology $ \tau_1 $ is considered **finer** than another topology $ \tau_2 $ if $ \tau_1 $ contains all the open sets of $ \tau_2 $. It is **strictly finer** if it contains more open sets than $ \tau_2 $.The task is to demonstrate that the uniform topology has more open sets than the product topology, meaning it can capture more nuanced behavior of sequences or functions in this space.

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Prove that for any infinite index set J, the uniform topology on R° is strictly finer than the product topology on R'.

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

**Prove that for any infinite index set $ J $, the uniform topology on $ \mathbb{R}^J $ is strictly finer than the product topology on $ \mathbb{R}^J $.**

This statement involves comparing two different topologies on the product space $ \mathbb{R}^J $, where $ J $ is an infinite index set:

1. **Product Topology**: This topology is generated by the basis of sets that are products of open intervals in $\mathbb{R}$, with all but finitely many factors being the entire real line $\mathbb{R}$.

2. **Uniform Topology**: This topology corresponds to the convergence of functions from $ J $ to $\mathbb{R}$ uniformly, that is, where the difference between function values is small across the entire index set $ J $.

A topology $ \tau_1 $ is considered **finer** than another topology $ \tau_2 $ if $ \tau_1 $ contains all the open sets of $ \tau_2 $. It is **strictly finer** if it contains more open sets than $ \tau_2 $.

The task is to demonstrate that the uniform topology has more open sets than the product topology, meaning it can capture more nuanced behavior of sequences or functions in this space.

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