1199: For Topological Group, Closed Subset Is Intersection of Subset Multiplied by Elements of Neighborhoods Basis at 1

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description/proof of that for topological group, closed subset is intersection of subset multiplied by elements of neighborhoods basis at $1$

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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: Structured Description
• 2: Proof

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Starting Context

• The reader knows a definition of topological group.
• The reader knows a definition of neighborhoods basis at point on topological space.
• The reader knows a definition of closed subset of topological space.
• The reader admits the proposition that for any topological group, any neighborhoods basis at $1$ satisfies these properties and each point multiplied by the neighborhoods basis at $1$ is a neighborhoods basis at the point.
• The reader admits the proposition that for any group with any topology with any continuous operations (especially, topological group), the set of the symmetric neighborhoods of $1$ is a neighborhood basis at $1$.
• The reader admits the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.

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Target Context

• The reader will have a description and a proof of the proposition that for any topological group, any closed subset is the intersection of the subset multiplied by the elements of any neighborhoods basis at $1$.

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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: Structured Description

Here is the rules of Structured Description.

Entities:
$G$: $\in \{\text{ the topological groups }\}$
$B_1$: $\in \{\text{ the neighborhoods bases at }1\text{ on }G\}$
$C$: $\in \{\text{ the closed subsets of }G\}$
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Statements:
$C={\cap }_{S_1\in B_1}{S}_{1}C={\cap }_{S_1\in B_1}C{S}_1$
//

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2: Proof

Whole Strategy: Step 1: see that $C\subseteq {\cap }_{S_1\in B_1}{S}_{1}C$; Step 2: see that ${\cap }_{S_1\in B_1}{S}_{1}C\subseteq C$, by seeing that for each $g\in {\cap }_{S_1\in B_1}{S}_{1}C$, $g\in C$ or $g$ is an accumulation point of $C$; Step 3: see that $C\subseteq {\cap }_{S_1\in B_1}C{S}_1$; Step 4: see that ${\cap }_{S_1\in B_1}C{S}_1\subseteq C$, by seeing that for each $g\in {\cap }_{S_1\in B_1}C{S}_1$, $g\in C$ or $g$ is an accumulation point of $C$.

Step 1:

Let us see that $C\subseteq {\cap }_{S_1\in B_1}{S}_{1}C$.

For each $c\in C$, for each $S_1\in B_1$, $c\in S_{1}C$, because $1\in S_1$, so, $c\in {\cap }_{S_1\in B_1}{S}_{1}C$.

Step 2:

Let us see that ${\cap }_{S_1\in B_1}{S}_{1}C\subseteq C$.

Let $g\in {\cap }_{S_1\in B_1}{S}_{1}C$ be any.

Let $N_g\subseteq G$ be any neighborhood of $g$.

As $g{B}_1$ is a neighborhood basis at $g$, by the proposition that for any topological group, any neighborhoods basis at $1$ satisfies these properties and each point multiplied by the neighborhoods basis at $1$ is a neighborhoods basis at the point, there is an $S_1\in B_1$ such that $g{S}_1\subseteq N_g$.

There is an $S_1^{\prime }\in B_1$ such that $S_1^{\prime }\subseteq g{S}_1{g}^{-1}$, by 4) in the proposition that for any topological group, any neighborhoods basis at $1$ satisfies these properties and each point multiplied by the neighborhoods basis at $1$ is a neighborhoods basis at the point.

There is a symmetric neighborhood of $1$, $S_1^{″}\subseteq G$, such that $S_1^{″}\subseteq S_1^{\prime }$, by the proposition that for any group with any topology with any continuous operations (especially, topological group), the set of the symmetric neighborhoods of $1$ is a neighborhood basis at $1$.

So, $S_1^{″}\subseteq g{S}_1{g}^{-1}$, so, $S_1^{″}g\subseteq g{S}_1{g}^{-1}g=g{S}_1$.

There is an $S_1^{‴}\in B_1$ such that $S_1^{‴}\subseteq S_1^{″}$.

$g\in S_1^{‴}C\subseteq S_1^{″}C$, because $g\in {\cap }_{S_1\in B_1}{S}_{1}C$.

That means that $g=sc$ where $s\in S_1^{″}$ and $c\in C$.

As $S_1^{″}g\subseteq g{S}_1$, $S_1^{″}sc\subseteq g{S}_1$, but as $S_1^{″}$ is symmetric, $s^{-1}\in S_1^{″}$, so, $c=s^{-1}sc\in g{S}_1\subseteq N_g$, which means that $g\in C$ or $g$ is an accumulation point of $C$, which means that $g$ is in the closure of $C$, by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset, so, $g\in \bar{C}$.

But as $C$ is closed, $\bar{C}=C$, so, $g\in \bar{C}=C$.

Step 3:

Let us see that $C\subseteq {\cap }_{S_1\in B_1}C{S}_1$.

As is expected, the logic is parallel to Step 1.

For each $c\in C$, for each $S_1\in B_1$, $c\in C{S}_1$, because $1\in S_1$, so, $c\in {\cap }_{S_1\in B_1}C{S}_1$.

Step 4:

Let us see that ${\cap }_{S_1\in B_1}C{S}_1\subseteq C$.

As is expected, the logic is parallel to Step 2, in fact, this is simpler: $S_1^{\prime }$ is not necessary as it is introduced in order to deal with the $S_{1}C$ order.

Let $g\in {\cap }_{S_1\in B_1}C{S}_1$ be any.

Let $N_g\subseteq G$ be any neighborhood of $g$.

As $g{B}_1$ is a neighborhood basis at $g$, by the proposition that for any topological group, any neighborhoods basis at $1$ satisfies these properties and each point multiplied by the neighborhoods basis at $1$ is a neighborhoods basis at the point, there is an $S_1\in B_1$ such that $g{S}_1\subseteq N_g$.

There is a symmetric neighborhood of $1$, $S_1^{″}\subseteq G$, such that $S_1^{″}\subseteq S_1$, by the proposition that for any group with any topology with any continuous operations (especially, topological group), the set of the symmetric neighborhoods of $1$ is a neighborhood basis at $1$.

There is an $S_1^{‴}\in B_1$ such that $S_1^{‴}\subseteq S_1^{″}$.

$g\in C{S}_1^{‴}\subseteq C{S}_1^{″}$, because $g\in {\cap }_{S_1\in B_1}C{S}_1$.

That means that $g=cs$ where $s\in S_1^{″}$ and $c\in C$.

As $S_1^{″}\subseteq S_1$, $g{S}_1^{″}=cs{S}_1^{″}\subseteq g{S}_1$, but as $S_1^{″}$ is symmetric, $s^{-1}\in S_1^{″}$, so, $c=cs{s}^{-1}\in g{S}_1\subseteq N_g$, which means that $g\in C$ or $g$ is an accumulation point of $C$, which means that $g$ is in the closure of $C$, by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset, so, $g\in \bar{C}$.

But as $C$ is closed, $\bar{C}=C$, so, $g\in \bar{C}=C$.

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References

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