definition of uniformly continuous map from group with topology with continuous operations (especially, topological group) into normed vectors space with induced topology
Topics
About: group
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of group.
- The reader knows a definition of topological space.
- The reader knows a definition of continuous, topological spaces map.
- The reader knows a definition of metric induced by norm on real or complex vectors space.
- The reader knows a definition of topology induced by metric.
Target Context
- The reader will have a definition of uniformly continuous map from group with topology with continuous operations (especially, topological group) into normed vectors space with induced topology.
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:
\( G\): \(\in \{\text{ the groups }\}\) with any topology such that the group operations are continuous
\( V\): \(\in \{\text{ the normed vectors spaces }\}\) with the induced topology induced by the metric induced by the norm
\(*f\): \(: G \to V\)
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Conditions:
\(\forall \epsilon \in \mathbb{R} \text{ such that } 0 \lt \epsilon (\exists U_1 \subseteq G \in \{\text{ the open neighborhoods of } 1 \} (\forall g_1, g_2 \in G \text{ such that } g_1 {g_2}^{-1} \in U_1 (\Vert f (g_1) - f (g_2) \Vert \lt \epsilon)))\)
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2: Note
\(g_1 {g_2}^{-1} \in U_1\) does not restrict \(g_1\) itself or \(g_2\) itself, because for each \(g_1 \in G\), \(g_2\) can be any in \({U_1}^{-1} g_1\), because then, \({g_2}^{-1} \in {g_1}^{-1} U_1\) and \(g_1 {g_2}^{-1} \in U_1\); for each \(g_2 \in G\), \(g_1\) can be any in \(U_1 g_2\), because then, \(g_1 {g_2}^{-1} \in U_1\).
\(f\) is inevitably continuous, because for each \(g_2 \in G\), \(U_1 g_2 \subseteq G\) is an open neighborhood of \(g_2\), by the proposition that for any group with any topology with any continuous operations (especially, any topological group), any element multiplied by any (open) neighborhood of \(1\) from left or right is a (open) neighborhood of the element, and for each \(g_1 \in U_1 g_2\), \(g_1 {g_2}^{-1} \in U_1\), so, \(\Vert f (g_1) - f (g_2) \Vert \lt \epsilon\), which means that \(f (U_1 g_2) \in B_{f (g_2), \epsilon}\).
