Prove the following statement: Let X and Y have bases B and C for their topologies, respectively. Then Dd (B x C |BEB,CEC) is a basis for the product topology on XXY.
Prove the following statement: Let X and Y have bases B and C for their topologies, respectively. Then Dd (B x C |BEB,CEC) is a basis for the product topology on XXY.
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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**Proving Product Topology Basis**
*Statement:*
Given two topological spaces \( X \) and \( Y \) with bases \( \mathcal{B} \) and \( \mathcal{C} \) for their respective topologies. We aim to prove that the collection
\[
\mathcal{D} \equiv \{ B \times C \mid B \in \mathcal{B}, C \in \mathcal{C} \}
\]
is a basis for the product topology on \( X \times Y \).
*Explanation:*
Consider two sets \( X \) and \( Y \). A topology on \( X \) and \( Y \) might be given by respective bases \( \mathcal{B} \) and \( \mathcal{C} \). For the product topology on \( X \times Y \), the set of all products of elements in \( \mathcal{B} \) and \( \mathcal{C} \) forms a basis.
**Steps to Prove:**
1. **Non-Empty Intersection Condition:**
Verify that for any point \( (x, y) \in X \times Y \) and any open set \( U \) containing \( (x, y) \), there exists a basis element \( B \times C \in \mathcal{D} \) such that \( (x, y) \in B \times C \subseteq U \).
2. **Basis Condition:**
Show that each open set in the product topology can be expressed as a union of basis elements.
This concept is fundamental in understanding product spaces within general topology, specifically how product topologies ensure structures from individual topological spaces are preserved.
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Transcribed Image Text:---
**Proving Product Topology Basis**
*Statement:*
Given two topological spaces \( X \) and \( Y \) with bases \( \mathcal{B} \) and \( \mathcal{C} \) for their respective topologies. We aim to prove that the collection
\[
\mathcal{D} \equiv \{ B \times C \mid B \in \mathcal{B}, C \in \mathcal{C} \}
\]
is a basis for the product topology on \( X \times Y \).
*Explanation:*
Consider two sets \( X \) and \( Y \). A topology on \( X \) and \( Y \) might be given by respective bases \( \mathcal{B} \) and \( \mathcal{C} \). For the product topology on \( X \times Y \), the set of all products of elements in \( \mathcal{B} \) and \( \mathcal{C} \) forms a basis.
**Steps to Prove:**
1. **Non-Empty Intersection Condition:**
Verify that for any point \( (x, y) \in X \times Y \) and any open set \( U \) containing \( (x, y) \), there exists a basis element \( B \times C \in \mathcal{D} \) such that \( (x, y) \in B \times C \subseteq U \).
2. **Basis Condition:**
Show that each open set in the product topology can be expressed as a union of basis elements.
This concept is fundamental in understanding product spaces within general topology, specifically how product topologies ensure structures from individual topological spaces are preserved.
---
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