description/proof of that topological space is regular iff for each point, 1-point subset is closed and set of closed neighborhoods of point is neighborhoods basis at point
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of regular topological space.
- The reader knows a definition of closed subset of topological space.
- The reader knows a definition of neighborhoods basis at point on topological space.
Target Context
- The reader will have a description and a proof of the proposition that any topological space is regular if and only if for each point of the space, the 1-point subset is closed and the set of the closed neighborhoods of the point is a neighborhoods basis at the point.
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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
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Statements:
\(T \in \{\text{ the regular topological spaces }\}\)
\(\iff\)
\(\forall t \in T (\{t\} \in \{\text{ the closed subsets of } T\} \land S_t := \{\text{ the closed neighborhoods of } t\} \in \{\text{ the neighborhoods bases at } t\})\)
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2: Note
When \(T\) is Hausdorff, each \(\{t\}\) is closed, by the proposition that for any Hausdorff topological space, any 1 point subset is closed, so, any Hausdorff \(T\) is regular if and only if \(S_t\) is a neighborhoods basis at \(t\).
3: Proof
Whole Strategy: Step 1: suppose that \(T\) is regular; Step 2: see that \(\{t\}\) is a closed subset and \(S_t\) is a neighborhoods basis at \(t\); Step 3: suppose that \(\{t\}\) is a closed subset and \(S_t\) is a neighborhoods basis at \(t\); Step 4: see that \(T\) is regular.
Step 1:
Let us suppose that \(T\) is regular.
Step 2:
Let \(t \in T\) be any.
\(\{t\}\) is a closed subset, by the definition of regular topological space.
Let \(N_t \subseteq T\) be any neighborhood of \(t\).
There is an open neighborhood of \(t\), \(U_t \subseteq T\), such that \(U_t \subseteq N_t\).
\(T \setminus U_t \subseteq T\) is closed and \(t \notin T \setminus U_t\).
As \(T\) is regular, there are an open neighborhood of \(t\), \(V_t \subseteq T\), and an open neighborhood of \(T \setminus U_t\), \(V_{T \setminus U_t} \subseteq T\), such that \(V_t \cap V_{T \setminus U_t} = \emptyset\).
Then, \(t \in T \setminus V_{T \setminus U_t} \subseteq U_t \subseteq N_t\), because as \(t \notin V_{T \setminus U_t}\), \(t \in T \setminus V_{T \setminus U_t}\), and \(T \setminus V_{T \setminus U_t} \subseteq T \setminus (T \setminus U_t) = U_t\).
And \(V_t \subseteq T \setminus V_{T \setminus U_t}\).
So, \(T \setminus V_{T \setminus U_t} \in S_t\).
So, \(S_t\) is a neighborhoods basis at \(t\).
Step 3:
Let us suppose that for each \(t \in T\), \(\{t\}\) is a closed subset and \(S_t\) is a neighborhoods basis at \(t\).
Step 4:
Let \(t \in T\) be any.
Let \(C \subseteq T\) be any closed subset such that \(t \notin C\).
\(T \setminus C \subseteq T\) is an open neighborhood of \(t\), because \(t \in T \setminus C\).
There is a \(C_t \in S_t\) such that \(C_t \subseteq T \setminus C\), by the supposition.
There is an open neighborhood of \(t\), \(U_t \subseteq T\), such that \(U_t \subseteq C_t\).
\(T \setminus C_t \subseteq T\) is an open neighborhood of \(C\), because \(C \subseteq T \setminus C_t\), because for each \(c \in C\), \(c \notin T \setminus C\), so, \(c \notin C_t\), so, \(c \in T \setminus C_t\).
\(U_t \cap (T \setminus C_t) = \emptyset\), because for each \(u \in U_t\), \(u \in C_t\), so, \(u \notin T \setminus C_t\).
So, \(T\) is regular.
