description/proof of that for continuous map from group with topology with continuous operations (especially, topological group) into normed vectors space with induced topology, if preimage of nonzeros is contained in compact subset of domain, map is uniformly-continuous
Topics
About: group
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of uniformly-continuous map from group with topology with continuous operations (especially, topological group) into normed vectors space with induced topology.
- The reader knows a definition of continuous, topological spaces map.
- The reader admits the proposition that for any group with any topology with any continuous operations (especially, topological group), any element, and any neighborhood of the element, there is a symmetric neighborhood of \(1\) such that the element multiplied from left by the neighborhood of \(1\) and multiplied from right by the inverse of the neighborhood of \(1\) is contained in the neighborhood of the element.
- The reader admits the proposition that for any group with any topology with any continuous operations (especially, topological group), any neighborhood of \(1\), and any positive natural number, there is a symmetric neighborhood of \(1\) whose power to the natural number is contained in the neighborhood.
- The reader admits 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.
- The reader admits the proposition that for any topological space and its any point, the intersection of any finite number of neighborhoods of the point is a neighborhood of the point.
- The reader admits the proposition that for any group, the intersection of any symmetric subsets is symmetric.
Target Context
- The reader will have a description and a proof of the proposition that for any continuous map from any group with topology with continuous operations (especially, topological group) into any normed vectors space with the induced topology, if the preimage of the nonzeros is contained in any compact subset of the domain, the map is uniformly-continuous.
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
\(f\): \(: G \to V\), \(\in \{\text{ the continuous maps }\}\)
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Statements:
\(\exists K \in \{\text{ the compact subsets of } G\} (f^{-1} (V \setminus \{0\}) \subseteq K)\)
\(\implies\)
\(f \in \{\text{ the uniformly-continuous maps }\}\)
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2: Proof
Whole Strategy: Step 1: for each \(k \in K\), take an open neighborhood of \(k\), \(U_k\), such that \(f (U_k) \subseteq B_{f (k), \epsilon / 2}\); Step 2: take a symmetric neighborhood of \(1\), \(W_{1, k}\), such that \(W_{1, k} W_{1, k} k \subseteq U_k\); Step 3: take a finite cover of \(K\), \(\{W_{1, k_j} k_j \vert j \in J\}\); Step 4: take \(W_1 := \cap_{j \in J} W_{1, k_j}\); Step 5: see that for each \(g, g' \in G\) such that \(g g'^{-1} \in W_1\), \(\Vert f (g) - f (g') \Vert \lt \epsilon\).
Step 1:
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
Let \(k \in K\) be any.
There is an open neighborhood of \(k\), \(U_k \subseteq G\), such that \(f (U_k) \subseteq B_{f (k), \epsilon / 2}\), because \(f\) is continuous.
Step 2:
There is a symmetric neighborhood of \(1\), \(W'_{1, k}\), such that \(W'_{1, k} k {W'_{1, k}}^{-1} \subseteq U_k\), by the proposition that for any group with any topology with any continuous operations (especially, topological group), any element, and any neighborhood of the element, there is a symmetric neighborhood of \(1\) such that the element multiplied from left by the neighborhood of \(1\) and multiplied from right by the inverse of the neighborhood of \(1\) is contained in the neighborhood of the element, then, \(W'_{1, k} k \subseteq W'_{1, k} k {W'_{1, k}}^{-1} \subseteq U_k\).
There is a symmetric neighborhood of \(1\), \(W_{1, k}\), such that \(W_{1, k} W_{1, k} \subseteq W'_{1, k}\), by the proposition that for any group with any topology with any continuous operations (especially, topological group), any neighborhood of \(1\), and any positive natural number, there is a symmetric neighborhood of \(1\) whose power to the natural number is contained in the neighborhood.
So, \(W_{1, k} W_{1, k} k \subseteq U_k\), so, \(W_{1, k} W_{1, k} \subseteq U_k k^{-1}\).
Also, \(W_{1, k} k \subseteq U_k\), because \(W_{1, k} \subseteq W_{1, k} W_{1, k}\), because for each \(w \in W_{1, k}\), \(w = w 1 \in W_{1, k} W_{1, k}\), so, \(W_{1, k} k \subseteq W_{1, k} W_{1, k} k\).
Step 3:
Each \(W_{1, k} k \subseteq G\) is a neighborhood of \(k\), 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, so, there is an open neighborhood of \(k\), \(V_k \subseteq G\), such that \(k \in V_k \subseteq W_{1, k} k\).
\(\{V_k \vert k \in K\}\) covers \(K\), so, \(\{V_k \vert k \in K\}\) is an open cover of \(K\).
As \(K\) is compact, there is a finite subcover, \(\{V_{k_j} \vert j \in J\}\).
\(\{W_{1, k_j} k_j \vert j \in J\}\) is a finite cover of \(K\).
Step 4:
Let us take \(W_1 := \cap_{j \in J} W_{1, k_j} \subseteq G\), which is a symmetric neighborhood of \(1\), by the proposition that for any topological space and its any point, the intersection of any finite number of neighborhoods of the point is a neighborhood of the point and the proposition that for any group, the intersection of any symmetric subsets is symmetric.
Step 5:
Let \(g, g' \in G\) be any such that \(g g'^{-1} \in W_1\).
When \(g, g' \notin K\), \(\Vert f (g) - f (g') \Vert = \Vert 0 - 0 \Vert = 0 \lt \epsilon\).
Let us suppose that \(g' \in K\).
\(g' \in W_{1, k_j} k_j \subseteq U_{k_j}\) for a \(j \in J\).
So, \(g' {k_j}^{-1} \in W_{1, k_j}\).
\(g g'^{-1} \in W_{1, k_j}\), because \(g g'^{-1} \in W_1 \subseteq W_{1, k_j}\).
\(g {k_j}^{-1} = (g g'^{-1}) (g' {k_j}^{-1}) \in W_{1, k_j} W_{1, k_j} \subseteq U_{k_j} {k_j}^{-1}\).
So, \(g \in U_{k_j}\).
As \(g, g' \in U_{k_j}\), \(\Vert f (g) - f (g') \Vert = \Vert f (g) - f (k_j) + f (k_j) - f (g') \Vert \le \Vert f (g) - f (k_j) \Vert + \Vert f (k_j) - f (g') \Vert \lt \epsilon / 2 + \epsilon / 2 = \epsilon\).
Let us suppose that \(g \in K\).
\(g \in W_{1, k_j} k_j \subseteq U_{k_j}\) for a \(j \in J\).
So, \(g {k_j}^{-1} \in W_{1, k_j}\).
\(g' g^{-1} \in W_{1, k_j}\), because \(g' g^{-1} = (g g'^{-1})^{-1} \in {W_1}^{-1} \subseteq W_{1, k_j}^{-1} = W_{1, k_j}\).
\(g' {k_j}^{-1} = (g' g^{-1}) (g {k_j}^{-1}) \in W_{1, k_j} W_{1, k_j} \subseteq U_{k_j} {k_j}^{-1}\).
So, \(g' \in U_{k_j}\).
As \(g, g' \in U_{k_j}\), \(\Vert f (g) - f (g') \Vert = \Vert f (g) - f (k_j) + f (k_j) - f (g') \Vert \le \Vert f (g) - f (k_j) \Vert + \Vert f (k_j) - f (g') \Vert \lt \epsilon / 2 + \epsilon / 2 = \epsilon\).
So, anyway, \(\Vert f (g) - f (g') \Vert \lt \epsilon\).
