description/proof of that for group with topology with continuous operations (especially, topological group), finite multiplication map is continuous
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 map.
- The reader admits the proposition that for any product topological space, any projection is continuous.
- The reader admits the proposition that any map from any topological space into any finite product topological space is continuous if and only if all the component maps are continuous.
- The reader admits the proposition that for any maps between any arbitrary subspaces of any topological spaces continuous at any corresponding points, the composition is continuous at the point.
Target Context
- The reader will have a description and a proof of the proposition that for any group with any topology with any continuous operations (especially, topological group), any finite multiplication map is 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
\(n\): \(\in \mathbb{N} \setminus \{0\}\)
\(f_n\): \(: G \times ... \times G \to G, (g_1, ..., g_n) \mapsto g_1 ... g_n\)
//
Statements:
\(f_n \in \{\text{ the continuous maps }\}\)
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2: Proof
Whole Strategy: prove it inductively; Step 1: see that it holds for \(n = 1\) and \(n = 2\); Step 2: suppose that it holds for any \(n = n' - 1\), and see that it holds for \(n = n'\); Step 3: conclude the proposition.
Step 1:
\(f_1\) is continuous, because it is the identity map.
\(f_2\) is continuous, because it it the multiplication map.
Step 2:
Let us suppose that for any \(n = n' - 1\), \(f_{n' - 1}\) is continuous.
Let us see that \(f_{n'}\) is continuous.
Let \(\pi_{1, ..., n' - 1}: G \times ... \times G \to G \times ... \times G, (g_1, ..., g_{n'}) \mapsto (g_1, ..., g_{n' - 1})\) be the projection, which is continuous, by the proposition that for any product topological space, any projection is continuous.
\(f_{n' - 1} \circ \pi_{1, ..., n' - 1}: G \times ... \times G \to G \times ... \times G \to G, (g_1, ..., g_{n'}) \mapsto (g_1, ..., g_{n' - 1}) \mapsto g_1 ... g_{n' - 1}\) is continuous, as a composition of continuous maps, by the proposition that for any maps between any arbitrary subspaces of any topological spaces continuous at any corresponding points, the composition is continuous at the point.
Let \(\pi_{n'}: G \times ... \times G \to G, (g_1, ..., g_{n'}) \mapsto g_{n'}\) be the projection, which is continuous, by the proposition that for any product topological space, any projection is continuous.
Let us take \(f': G \times ... \times G \to G \times G, g = (g_1, ..., g_{n'}) \mapsto (f_{n' - 1} \circ \pi_{1, ..., n' - 1} (g), \pi_{n'} (g)) = (g_1 ... g_{n' - 1}, g_{n'})\), which is continuous, by the proposition that any map from any topological space into any finite product topological space is continuous if and only if all the component maps are continuous.
\(f_{n'} : G \times ... \times G \to G \times G \to G, g = (g_1, ..., g_{n'}) \mapsto (g_1 ... g_{n' - 1}, g_{n'}) \mapsto g_1 ... g_{n' - 1} g_{n'} = f_2 \circ f'\), which is continuous as a composition of continuous maps.
Step 3:
So, by the induction principle, for each \(n \in \mathbb{N} \setminus \{0\}\), \(f_n\) is continuous.
