description/proof of that paracompact Hausdorff topological space is normal
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of paracompact topological space.
- The reader knows a definition of Hausdorff topological space.
- The reader knows a definition of normal topological space.
- The reader admits the proposition that for any locally finite set of subsets of any topological space, the closure of the union of the subsets is the union of the closures of the subsets.
- The reader admits the proposition that for any Hausdorff topological space, any 1 point subset is closed.
Target Context
- The reader will have a description and a proof of the proposition that any paracompact Hausdorff topological space is normal.
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:
\(T\): \(= \{\text{ the paracompact Hausdorff topological spaces }\}\)
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Statements:
\(T \in \{\text{ the normal topological spaces }\}\)
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2: Proof
Whole Strategy: Step 1: see that \(T\) is regular; Step 2: see that \(T\) is normal.
Step 1:
Step 1 Strategy: Step 1-1: see that each \(1\)-point subset of \(T\) is closed; Step 1-2: for each \(t \notin C\), take an open cover of \(T\), \(\{U_c \vert c \in C\} \cup \{T \setminus C\}\), where \(U_{t, c} \cap U_c = \emptyset\), where \(U_{t, c}\) is an open neighborhood of \(t\), take a locally finite refinement, \(\{V_l \vert l \in L_1\} \cup \{V_l \vert l \in L_2\}\) where \(V_l \subseteq U_c\) for \(l \in L_1\) and \(V_l \subseteq T \setminus C\) for \(l \in L_2\); Step 1-3: see that \(V_C := \cup_{l \in L_1} V_l\) contains \(C\); Step 1-4: see that \(t \in T \setminus \overline{V_C}\).
Step 1-1:
Each \(1\)-point subset of \(T\) is closed, by the proposition that for any Hausdorff topological space, any 1 point subset is closed.
Step 1-2:
Let \(t \in T\) be any.
Let \(C \subseteq T\) be any closed subset such that \(t \notin C\).
For each \(c \in C\), there are an open neighborhood of \(t\), \(U_{t, c} \subseteq T\), and an open neighborhood of \(c\), \(U_c \subseteq T\), such that \(U_{t, c} \cap U_c = \emptyset\), because \(T\) is Hausdorff.
\(\{U_c \vert c \in C\} \cup \{T \setminus C\}\) is an open cover of \(T\), because for each \(t \in T\), \(t \in C\) or \(t \in T \setminus C\), and when \(t \in C\), \(t \in U_t\), and when \(t \in T \setminus C\), \(t \in T \setminus C\).
There is a locally finite refinement of \(\{U_c \vert c \in C\} \cup \{T \setminus C\}\), \(\{V_l \vert l \in L_1\} \cup \{V_l \vert l \in L_2\}\) where \(V_l \subseteq U_c\) for each \(l \in L_1\) and \(V_l \subseteq T \setminus C\) for each \(l \in L_2\): it is the partition of the locally finite refinement such that each \(V_l\) such that \(V_l \subseteq U_c\) is put in \(L_1\) and the others are put in \(L_2\).
Step 1-3:
Let us define \(V_C := \cup_{l \in L_1} V_l\).
\(C \subseteq V_C\), because for each \(c \in C\), \(c \in V_l\) for an \(l \in L_1 \cup L_2\), but if \(l \in L_2\), \(c \in V_l \subseteq T \setminus C\), a contradiction against \(c \in C\), so, \(l \in L_1\), so, \(c \in \cup_{l \in L_1} V_l\).
Step 1-4:
\(\overline{V_C} = \overline{\cup_{l \in L_1} V_l} = \cup_{l \in L_1} \overline{V_l}\), by the proposition that for any locally finite set of subsets of any topological space, the closure of the union of the subsets is the union of the closures of the subsets.
For each \(l \in L_1\), \(t \notin \overline{V_l}\), because while \(V_l \subseteq U_c\) for a \(c \in C\), \(t \in U_{t, c}\), but \(U_c \subseteq T \setminus U_{t, c}\), so, \(V_l \subseteq U_c \subseteq T \setminus U_{t, c}\), so, \(\overline{V_l} \subseteq T \setminus U_{t, c}\), so, as \(t \notin T \setminus U_{t, c}\), \(t \notin \overline{V_l}\).
So, \(t \notin \cup_{l \in L_1} \overline{V_l} = \overline{V_C}\).
So, \(t \in T \setminus \overline{V_C}\).
\(T \setminus \overline{V_C}\) is an open neighborhood of \(t\).
\((T \setminus \overline{V_C}) \cap V_C = \emptyset\).
So, \(T\) is regular.
Step 2:
Step 2 Strategy: Step 2-1: for each \(C_1 \cap C_2 = \emptyset\), take an open cover of \(T\), \(\{U_{c_2} \vert c_2 \in C_2\} \cup \{T \setminus C_2\}\) where \(U_{c_2} \cap U_{C_1, c_2} = \emptyset\) where \(U_{C_1, c_2}\) is an open neighborhood of \(C_1\); Step 2-2: take a locally finite refinement, \(\{V_l \vert l \in L_1\} \cup \{V_l \vert l \in L_2\}\) where \(V_l \subseteq U_{c_2}\) for \(l \in L_1\) and \(V_l \subseteq T \setminus C_2\) for \(l \in L_2\); Step 2-3: see that \(U_{C_2} := \cup_{l \in L_1} V_l\) covers \(C_2\); Step 2-4: see that \(C_1 \subseteq T \setminus \overline{V_{C_2}}\).
Step 2-1:
Let \(C_1, C_2 \subseteq T\) be any closed subsets such that \(C_1 \cap C_2 = \emptyset\).
Let \(c_2 \in C_2\) be any.
There are an open neighborhood of \(c_2\), \(U_{c_2} \subseteq T\), and an open neighborhood of \(C_1\), \(U_{C_1, c_2} \subseteq T\), such that \(U_{c_2} \cap U_{C_1, c_2} = \emptyset\), by Step 1.
\(\{U_{c_2} \vert c_2 \in C_2\} \cup \{T \setminus C_2\}\) is an open cover of \(T\), because for each \(t \in T\), \(t \in C_2\) or \(t \in T \setminus C_2\), and when \(t \in C_2\), \(t \in U_t\), and when \(t \in T \setminus C_2\), \(t \in T \setminus C_2\).
Step 2-2:
There is a locally finite refinement, \(\{V_l \vert l \in L_1\} \cup \{V_l \vert l \in L_2\}\) where \(V_l \subseteq U_{c_2}\) for each \(l \in L_1\) and \(V_l \subseteq T \setminus C_2\) for each \(l \in L_2\): it is the partition of the refinement such that each \(V_l\) such that \(V_l \subseteq U_{c_2}\) for a \(U_{c_2}\) is put into \(L_1\) and the others are put into \(L_2\).
Step 2-3:
Let us define \(U_{C_2} := \cup_{l \in L_1} V_l\).
\(C_2 \subseteq U_{C_2}\), because for each \(c_2 \in C_2\), \(c_2 \in V_l\) for a \(l \in L_1 \cup L_2\), but if \(l \in L_2\), \(V_l \subseteq T \setminus C_2\), so, \(c_2 \in V_l \subseteq T \setminus C_2\), a contradiction against \(c_2 \in C_2\), so, \(l \in L_1\), so, \(c_2 \in \cup_{l \in L_1} V_l = U_{C_2}\).
Step 2-4:
\(\overline{U_{C_2}} = \overline{\cup_{l \in L_1} V_l} = \cup_{l \in L_1} \overline{V_l}\), by the proposition that for any locally finite set of subsets of any topological space, the closure of the union of the subsets is the union of the closures of the subsets.
For each \(l \in L_1\), \(V_l \subseteq U_{c_2}\) for a \(c_2 \in C_2\), but \(U_{c_2} \subseteq T \setminus U_{C_1, c_2}\), so, \(V_l \subseteq U_{c_2} \subseteq T \setminus U_{C_1, c_2}\), so, \(\overline{V_l} \subseteq T \setminus U_{C_1, c_2} \subseteq T \setminus C_1\).
So, \(\overline{U_{C_2}} = \cup_{l \in L_1} \overline{V_l} \subseteq T \setminus C_1\).
So, \(C_1 \subseteq T \setminus \overline{U_{C_2}}\).
\(T \setminus \overline{U_{C_2}}\) is an open neighborhood of \(C_1\).
\((T \setminus \overline{V_{C_2}}) \cap V_{C_2} = \emptyset\).
As we already know that each \(1\)-point subset of \(T\) is closed, \(T\) is normal.
