description/proof of that for finite-product metric space and open ball, there is product of open balls contained in open ball in which (product) open ball is contained
Topics
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of finite-product metric space.
Target Context
- The reader will have a description and a proof of the proposition that for any finite-product metric space and any open ball, there is the product of some open balls contained in the open ball in which (the product) an open ball is contained.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(J\): \(\in \{\text{ the finite index sets }\}\)
\(\{M_j \vert j \in J\}\): \(M_j \in \{\text{ the metric spaces }\}\) with any metric, \(dist_j\)
\(\times_{j \in J} M_j\): \(= \text{ the finite-product metric space }\) with the finite-product metric, \(dist\)
\(m\): \(\in \times_{j \in J} M_j\)
\(B_{m, \epsilon}\): \(\in \{\text{ the open balls around } m\}\)
\(\beta\): \(= \epsilon / \sqrt{\vert J \vert}\)
//
Statements:
\(B_{m, \beta} \subseteq \times_{j \in J} B_{m^j, \beta} \subseteq B_{m, \epsilon}\)
//
2: Proof
Whole Strategy: Step 1: see that \(\times_{j \in J} B_{m^j, \beta} \subseteq B_{m, \epsilon}\); Step 2: see that \(B_{m, \beta} \subseteq \times_{j \in J} B_{m^j, \beta}\).
Step 1:
Let us see that \(\times_{j \in J} B_{m^j, \beta} \subseteq B_{m, \epsilon}\).
Let \(p \in \times_{j \in J} B_{m^j, \beta}\) be any.
\(dist (m, p) = \sqrt{\sum_{j \in J} dist_j (m^j, p^j)^2}\).
As \(p^j \in B_{m^j, \beta}\), \(dist_j (m^j, p^j) \lt \beta\).
So, \(\sqrt{\sum_{j \in J} dist_j (m^j, p^j)^2} \lt \sqrt{\sum_{j \in J} \beta^2} = \sqrt{\sum_{j \in J} \epsilon^2 / \vert J \vert} = \epsilon\).
So, \(p \in B_{m, \epsilon}\).
Step 2:
Let us see that \(B_{m, \beta} \subseteq \times_{j \in J} B_{m^j, \beta}\).
Let \(p \in B_{m, \beta}\) be any.
\(dist (m, p) \lt \beta\).
\(dist (m, p) = \sqrt{\sum_{j \in J} dist_j (m^j, p^j)^2}\).
For each \(j \in J\), \(dist_j (m^j, p^j)^2 \le \sum_{j \in J} dist_j (m^j, p^j)^2\), so, \(dist_j (m^j, p^j) \le \sqrt{\sum_{j \in J} dist_j (m^j, p^j)^2} \lt \beta\).
So, \(p^j \in B_{m^j, \beta}\).
So, \(p \in \times_{j \in J} B_{m^j, \beta}\).
