description/proof of that topological space is normal iff for each open subset, for each closed subset contained in open subset, there is open subset s.t. it contains closed subset and its closure is contained in open subset
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of normal topological space.
- The reader knows a definition of closure of subset of topological space.
- The reader admits the proposition that for any set and any \(2\) disjoint subsets, the 1st subset is contained in the complement of the 2nd subset.
- The reader admits the proposition that for any set and any \(2\) subsets, if the complement of the 1st subset is contained in the 2nd subset, the complement of the 2nd subset is contained in the 1st subset.
Target Context
- The reader will have a description and a proof of the proposition that any topological space is normal if and only if for each open subset, for each closed subset contained in the open subset, there is an open subset such that it contains the closed subset and its closure is contained in the open subset.
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:
\(T \in \{\text{ the normal topological spaces }\}\)
\(\iff\)
\(\forall U \in \{\text{ the open subsets of } T\} (\forall C \in \{\text{ the closed subsets of } T\} \text{ such that } C \subseteq U (\exists V \in \{\text{ the open subsets of } T\} (C \subseteq V \subseteq \overline{V} \subseteq U)))\)
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2: Proof
Whole Strategy: Step 1: suppose that \(T\) is normal; Step 2: take \(U_1, U_2\) such that \(C \subseteq U_1\), \(T \setminus U \subseteq U_2\), and \(U_1 \cap U_2 = \emptyset\), and see that \(V := U_1\) will do; Step 3: suppose that there is a \(V\); Step 4: take \(C_1, C_2\) such that \(C_1 \cap C_2 = \emptyset\), and see that \(V\) for \(C_1 \subseteq T \setminus C_2\) and \(T \setminus \overline{V}\) will do.
Step 1:
Let us suppose that \(T\) is normal.
Step 2:
\(T \setminus U \subseteq T\) is closed.
\(C \cap (T \setminus U) = \emptyset\), because for each \(c \in C\), \(c \in U\), so, \(c \notin T \setminus U\).
As \(T\) is normal, there are some open subsets, \(U_1, U_2 \subseteq T\), such that \(C \subseteq U_1\), \(T \setminus U \subseteq U_2\), and \(U_1 \cap U_2 = \emptyset\).
\(U_1 \subseteq U\), because as \(U_1 \cap U_2 = \emptyset\) and \(T \setminus U \subseteq U_2\), \(U_1 \cap (T \setminus U) = \emptyset\), and by the proposition that for any set and any \(2\) disjoint subsets, the 1st subset is contained in the complement of the 2nd subset, \(U_1 \subseteq (T \setminus (T \setminus U)) = U\).
\(U_1 \subseteq T \setminus U_2\), by the proposition that for any set and any \(2\) disjoint subsets, the 1st subset is contained in the complement of the 2nd subset, because \(U_1 \cap U_2 = \emptyset\).
So, \(\overline{U_1} \subseteq T \setminus U_2\), because the closure is the intersection of the closed subsets that contain \(U_1\) while \(T \setminus U_2\) is one of such closed subsets.
\(T \setminus U_2 \subseteq U\), by the proposition that for any set and any \(2\) subsets, if the complement of the 1st subset is contained in the 2nd subset, the complement of the 2nd subset is contained in the 1st subset, because \(T \setminus U \subseteq U_2\).
So, \(C \subseteq U_1 \subseteq \overline{U_1} \subseteq U\).
So, \(V := U_1\) will do.
Step 3:
Let us suppose that there is a \(V\) for each \(U\) and each \(C\).
Step 4:
Let \(C_1, C_2 \subseteq T\) be any closed subsets such that \(C_1 \cap C_2 = \emptyset\).
Let \(C := C_1\) and \(U := T \setminus C_2\).
\(U \subseteq T\) is open, and \(C \subseteq U\), by the proposition that for any set and any \(2\) disjoint subsets, the 1st subset is contained in the complement of the 2nd subset, because \(C_1 \cap C_2 = \emptyset\).
So, there is an open \(V \subseteq T\) such that \(C \subseteq V \subseteq \overline{V} \subseteq U\).
\(C_2 \subseteq T \setminus \overline{V}\), because for each \(c_2 \in C_2\), \(c_2 \notin U\), so, \(c_2 \notin \overline{V}\), so, \(c_2 \in T \setminus \overline{V}\).
\(T \setminus \overline{V} \subseteq T\) is open.
\(V \cap (T \setminus \overline{V}) = \emptyset\).
So, \(C_1 \subseteq V\), \(C_2 \subseteq T \setminus \overline{V}\), and \(V \cap (T \setminus \overline{V}) = \emptyset\).
So, \(T\) is normal.
