description/proof of that for map from finite-product metric space into metric space continuous at point, induced map with set of components of domain fixed is continuous at point
Topics
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of finite-product metric space.
- The reader knows a definition of metric spaces map continuous at point.
- The reader admits 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.
Target Context
- The reader will have a description and a proof of the proposition that for any map from any finite-product metric space into any metric space continuous at any point, the induced map with any set of the components of the domain fixed is continuous at the point.
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, \(dist\)
\(M\): \(\in \{\text{ the metric spaces }\}\)
\(f'\): \(: \times_{j \in J'} M_j \to M\)
\(J\): \(\subseteq J'\)
\(\pi_1\): \(: \times_{j \in J'} M_j \to \times_{j \in J} M_j\), \(= \text{ the projection }\)
\(\pi_2\): \(: \times_{j \in J'} M_j \to \times_{j \in J' \setminus J} M_j\), \(= \text{ the projection }\)
\(m'\): \(\in \times_{j \in J'} M_j\)
\(g_{m'}\): \(: \times_{j \in J} M_j \to \times_{j \in J'} M_j, m \mapsto g_{m'} (m) \text{ such that } \pi_1 (g_{m'} (m)) = m \land \pi_2 (g_{m'} (m)) = \pi_2 (m')\)
\(f_{m'}\): \(: \times_{j \in J} M_j \to M, m \mapsto f' (g_{m'} (m))\)
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Statements:
\(f' \in \{\text{ the maps continuous at } m'\}\)
\(\implies\)
\(f_{m'} \in \{\text{ the maps continuous at } \pi_1 (m')\}\)
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2: Note
When \(f'\) is continuous, \(f_{m'}\) is continuous: for each \(m'' \in \times_{j \in J'} M_j\) such that \(\pi_2 (m') = \pi_2 (m'')\), \(f_{m'} = f_{m''}\) and \(\pi_1 (m'')\) s cover \(\times_{j \in J} M_j\).
3: Proof
Whole Strategy: Step 1: take any \(B_{f_{m'} (\pi_1 (m')), \epsilon}\) and a \(B_{m', \delta}\) such that \(f' (B_{m', \delta}) \subseteq B_{f_{m'} (\pi_1 (m')), \epsilon}\); Step 2: take a \(B_{\pi_1 (m'), \beta}\) such that \(g_{m'} (B_{\pi_1 (m'), \beta}) \subseteq B_{m', \delta}\); Step 3: see that \(f_{m'} (B_{\pi_1 (m'), \beta}) \subseteq B_{f_{m'} (\pi_1 (m')), \epsilon}\).
Step 1:
Let us take any \(B_{f_{m'} (\pi_1 (m')), \epsilon} \subseteq M\).
\(g_{m'} (\pi_1 (m')) = m'\), and \(f_{m'} (\pi_1 (m')) = f' (g_{m'} (\pi_1 (m'))) = f' (m')\).
So, \(B_{f_{m'} (\pi_1 (m')), \epsilon} = B_{f' (m'), \epsilon}\).
As \(f'\) is continuous at \(m'\), there is a \(B_{m', \delta} \subseteq \times_{j \in J'} M_j\) such that \(f' (B_{m', \delta}) \subseteq B_{f_{m'} (\pi_1 (m')), \epsilon}\).
Step 2:
There is \(\beta := \delta / \sqrt{\vert J' \vert}\) such that \(\times_{j \in J'} B_{m'^j, \beta} \subseteq B_{m', \delta}\), by 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.
Then, \(\times_{j \in J} B_{m'^j, \beta} \subseteq \times_{j \in J} M_j\) satisfies \(g_{m'} (\times_{j \in J} B_{m'^j, \beta}) \subseteq \times_{j \in J'} B_{m'^j, \beta}\), because for each \(p \in \times_{j \in J} B_{m'^j, \beta}\), \(\pi_1 (g_{m'} (p)) = p\), so, \(\pi_1 (g_{m'} (p))^j = p^j \in B_{m'^j, \beta}\), and \(\pi_2 (g_{m'} (p)) = \pi_2 (m')\), so, \(\pi_2 (g_{m'} (p))^j = \pi_2 (m')^j \in B_{m'^j, \beta}\).
But \(\times_{j \in J} B_{m'^j, \beta} = \times_{j \in J} B_{\pi_1 (m')^j, \beta}\).
\(B_{\pi_1 (m'), \beta} \subseteq \times_{j \in J} B_{\pi_1 (m')^j, \beta}\), by 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.
So, \(g_{m'} (B_{\pi_1 (m'), \beta}) \subseteq g_{m'} (\times_{j \in J} B_{\pi_1 (m')^j, \beta}) \subseteq \times_{j \in J'} B_{m'^j, \beta} \subseteq B_{m', \delta}\).
Step 3:
So, \(f_{m'} (B_{\pi_1 (m'), \beta}) = f' \circ g_{m'} (B_{\pi_1 (m'), \beta}) \subseteq f' (B_{m', \delta}) \subseteq B_{f_{m'} (\pi_1 (m')), \epsilon}\).
So, \(f_{m'}\) is continuous at \(\pi_1 (m')\).
