334: Closure of Subgroup of Topological Group Is Subgroup

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A description/proof of that closure of subgroup of topological group is subgroup

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Topics

About: topological group

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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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.

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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.

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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.

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Main Body

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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.

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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 $T_2$, ${N_p}^{-1}\cap T_2\ne \mathrm{\varnothing }$. So, there is a point, $p^{\prime }\in {N_p}^{-1}\cap T_2$, and $p^{\prime -1}\in N_p\cap T_2$, so, $N_p\cap T_2\ne \mathrm{\varnothing }$. 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, $f:T_1\times T_1\to T_1$, 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, $N_{⟨p_1,p_2⟩}\subseteq T_1\times T_1$, of $⟨p_1,p_2⟩$ such that $f(N_{⟨p_1,p_2⟩})\subseteq N_{p_1{p}_2}$ by the definition of continuous map. There is a open neighborhood, $N_{⟨p_1,p_2⟩}^{\prime }\subseteq N_{⟨p_1,p_2⟩}$, of $⟨p_1,p_2⟩$, and $N_{⟨p_1,p_2⟩}^{\prime }={\cup }_{\alpha \in A}{N}_{p_1-\alpha }^{\prime }\times N_{p_2-\alpha }^{\prime }$ where $A$ is a possibly uncountable indexes set and $N_{p_i-\alpha }^{\prime }\subseteq T_1$ is an open neighborhood of $p_i$. For an $\alpha \in A$, $⟨p_1,p_2⟩\in N_{p_1-\alpha }^{\prime }\times N_{p_2-\alpha }^{\prime }$ and $f(N_{p_1-\alpha }^{\prime }\times N_{p_2-\alpha }^{\prime })\subseteq N_{p_1{p}_2}$. Whether $p_i$ is a point of $T_2$ or an accumulation point of $T_2$, there are points, $p_1^{\prime }\in N_{p_1-\alpha }^{\prime }\cap T_2$ and $p_2^{\prime }\in N_{p_2-\alpha }^{\prime }\cap T_2$, and $f(⟨p_1^{\prime },p_2^{\prime }⟩)=p_1^{\prime }{p}_2^{\prime }\in N_{p_1{p}_2}\cap T_2$. So, $p_1{p}_2$ is a point of $T_2$ or an accumulation point of $T_2$.

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References

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