A description/proof of that closure of subgroup of topological group is subgroup
Topics
About: topological group
The table of contents of this article
Starting Context
- The reader knows a definition of topological group.
- The reader knows a definition of subgroup of topological group.
- The reader knows a definition of closure of subset.
- The reader admits the proposition that the inverse map on any topological group is a homeomorphism.
Target Context
- The reader will have a description and a proof of the proposition that the closure of any subgroup of any topological group is a subgroup.
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: Description
For any topological group, \(T_1\), and any subgroup, \(T_2 \subseteq T_1\), the closure, \(\overline{T_2}\), of \(T_2\) is a subgroup.
2: Proof
It is about \(\overline{T_2}^{-1} \subseteq \overline{T_2}\) and \(\overline{T_2} \overline{T_2} \subseteq \overline{T_2}\).
Let us prove that \(\overline{T_2}^{-1} \subseteq \overline{T_2}\). For any \(p \in \overline{T_2}^{-1}\), \(p \in \overline{T_2}\)? \(p^{-1} \in \overline{T_2}\). For any neighborhood, \(N_p \subseteq T_1\), of \(p\), \(p^{-1} \in {N_p}^{-1}\), but by the proposition that the inverse map on any topological group is a homeomorphism, \({N_p}^{-1}\) is a neighborhood of \(p^{-1}\) on \(T_1\). Whether \(p^{-1}\) is a point of \(T_2\) or an accumulation point of \(T_2\), \({N_p}^{-1} \cap T_2 \neq \emptyset\). So, there is a point, \(p' \in {N_p}^{-1} \cap T_2\), and \(p'^{-1} \in N_p \cap T_2\), so, \(N_p \cap T_2 \neq \emptyset\). So, \(p\) is a point of \(T_2\) or an accumulation point of \(T_2\).
Let us prove that \(\overline{T_2} \overline{T_2} \subseteq \overline{T_2}\). For any points, \(p_1, p_2 \in \overline{T_2}\), \(p_1 p_2 \in \overline{T_2}\)? As the multiplication map, \(f: T_1 \times T_1 \rightarrow T_1\), is continuous by the definition of topological group, for any neighborhood, \(N_{p_1 p_2} \subseteq T_1\), of \(p_1 p_2\), there is a neighborhood, \(N_{\langle p_1, p_2 \rangle } \subseteq T_1 \times T_1\), of \(\langle p_1, p_2 \rangle \) such that \(f (N_{\langle p_1, p_2 \rangle }) \subseteq N_{p_1 p_2}\) by the definition of continuous map. There is a open neighborhood, \(N'_{\langle p_1, p_2 \rangle } \subseteq N_{\langle p_1, p_2 \rangle }\), of \(\langle p_1, p_2 \rangle \), and \(N'_{\langle p_1, p_2 \rangle } = \cup_{\alpha \in A} N'_{p_1-\alpha} \times N'_{p_2-\alpha}\) where \(A\) is a possibly uncountable indexes set and \(N'_{p_i-\alpha} \subseteq T_1\) is an open neighborhood of \(p_i\). For an \(\alpha \in A\), \(\langle p_1, p_2 \rangle \in N'_{p_1-\alpha} \times N'_{p_2-\alpha}\) and \(f (N'_{p_1-\alpha} \times N'_{p_2-\alpha}) \subseteq N_{p_1 p_2}\). Whether \(p_i\) is a point of \(T_2\) or an accumulation point of \(T_2\), there are points, \(p'_1 \in N'_{p_1-\alpha} \cap T_2\) and \(p'_2 \in N'_{p_2-\alpha} \cap T_2\), and \(f (\langle p'_1, p'_2 \rangle ) = p'_1 p'_2 \in N_{p_1 p_2} \cap T_2\). So, \(p_1 p_2\) is a point of \(T_2\) or an accumulation point of \(T_2\).
