description/proof of that for topological group, closed subset is intersection of subset multiplied by elements of neighborhoods basis at \(1\)
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 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.
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 \(1\).
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:
\(G\): \(\in \{\text{ the topological groups }\}\)
\(B_1\): \(\in \{\text{ the neighborhoods bases at } 1 \text{ on } G\}\)
\(C\): \(\in \{\text{ the closed subsets of } G\}\)
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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 \(S'_1 \in B_1\) such that \(S'_1 \subseteq g S_1 g^{-1}\), 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 \(S''_1 \subseteq S'_1\), 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 \(g = s c\) where \(s \in S''_1\) and \(c \in C\).
As \(S''_1 g \subseteq g S_1\), \(S''_1 s c \subseteq g S_1\), but as \(S''_1\) is symmetric, \(s^{-1} \in S''_1\), so, \(c = s^{-1} s c \in g S_1 \subseteq N_g\), 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, \(g \in \overline{C}\).
But as \(C\) is closed, \(\overline{C} = C\), so, \(g \in \overline{C} = C\).
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: \(S'_1\) 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 \(g = c s\) where \(s \in S''_1\) and \(c \in C\).
As \(S''_1 \subseteq S_1\), \(g S''_1 = c s S''_1 \subseteq g S_1\), but as \(S''_1\) is symmetric, \(s^{-1} \in S''_1\), so, \(c = c s s^{-1} \in g S_1 \subseteq N_g\), 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, \(g \in \overline{C}\).
But as \(C\) is closed, \(\overline{C} = C\), so, \(g \in \overline{C} = C\).
