description/proof of for paracompact Hausdorff topological space, open cover of space, and locally finite refinement, there is partition of unity subordinate to refinement
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 partition of unity subordinate to open cover of topological space.
- The reader admits the proposition that any set with the ordering as containment is a partially-ordered set.
- The reader admits the proposition that any paracompact Hausdorff topological space is normal.
- The reader admits 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.
- The reader admits the proposition that for any normal topological space, any closed subset, and any open subset that contains the closed subset, there is a continuous map into any closed Interval that maps the closed subset to any boundary of the closed interval and the complement of the open subset to the other boundary (Urysohn's lemma).
- The reader admits the proposition that for any linearly-ordered set, any finite subset can be ordered in a line.
- The reader admits the Hausdorff maximal principle that any chain of any partially-ordered set is contained in a maximal chain.
- The reader admits the proposition that for any topological space and any set of continuous maps from the space into any Euclidean topological space such that the set of the preimages of nonzero is locally finite, the sum of the maps is continuous.
Target Context
- The reader will have a description and a proof of the proposition that for any paracompact Hausdorff topological space, any open cover of the space, and any locally finite refinement of the cover, there is a partition of unity subordinate to the refinement.
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:
\(T\): \(\in \{\text{ the paracompact Hausdorff topological spaces }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{U_j \in \{\text{ the open subsets of } T\} \vert j \in J\}\): such that \(T = \cup_{j \in J} U_j\)
\(L\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{V_l \vert l \in L\}\): \(\in \{\text{ the locally finite refinements of } \{U_j \vert j \in J\}\}\)
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Statements:
\(\exists \{\rho_l \vert l \in L\} \in \{\text{ the partitions of unity subordinate to } \{V_l \vert l \in L\}\}\)
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2: Note
For any \(\{U_j \vert j \in J\}\), there is a \(\{V_l \vert l \in L\}\), by the definition of paracompact topological space, and for any choice of \(\{V_l \vert l \in L\}\), there is a \(\{\rho_l \vert l \in L\}\), by this proposition.
There is a partition of unity subordinate to \(\{U_j \vert j \in J\}\), by the proposition that for any topological space and any open cover of the space, if there are any locally finite refinement of the open cover and any partition of unity subordinate to the refinement, there is a partition of unity subordinate to the original cover.
3: Proof
Whole Strategy: Step 1: take the sets, \(F := \{f: T \to [0, 1] \vert f \in \{\text{ the continuous maps }\}\}\), \(S'' := \cup_{L^` \subseteq L} \{g: L^` \to F\}\), \(S' := \{g \in S'' \vert \forall l \in Dom (g) (\overline{g (l)^{-1} ([0, 1] \setminus \{0\})} \subseteq V_l) \land T = (\cup_{l \in Dom (g)} g (l)^{-1} ([0, 1] \setminus \{0\})) \cup (\cup_{l \in L \setminus Dom (g)} V_l)\}\), and \(S := \{\{(l, g (l)) \vert l \in Dom (g)\} \vert g \in S'\}\); Step 2: take a maximal chain of \(S\), \(C\), and \(s := \cup C\); Step 3: see that \(s \in C\) with \(Dom (s) = L\); Step 4: take \(\rho := \sum_{l \in L} s (l)\) and \(\rho_l := s (l) / \rho\); Step 5: see that \(\{\rho_l \vert l \in L\}\) is a partition of unity subordinate to \(\{V_l \vert l \in L\}\).
Step 1:
Let us take \(F := \{f: T \to [0, 1] \vert f \in \{\text{ the continuous maps }\}\}\), which is a legitimate set.
Let us take \(S'' := \cup_{L^` \subseteq L} \{g: L^` \to F\}\), which is a legitimate set.
Let us take \(S' := \{g \in S'' \vert \forall l \in Dom (g) (\overline{g (l)^{-1} ([0, 1] \setminus \{0\})} \subseteq V_l) \land T = (\cup_{l \in Dom (g)} g (l)^{-1} ([0, 1] \setminus \{0\})) \cup (\cup_{l \in L \setminus Dom (g)} V_l)\}\), which is a legitimate set.
Let us take \(S := \{\{(l, g (l)) \vert l \in Dom (g)\} \vert g \in S'\}\), which is a legitimate set.
Note that \(S\) is essentially the same with \(S'\) with the difference that each map, \(g \in S'\), is expressed as the relation, \(\{(l, g (l)) \vert l \in Dom (g)\} \subseteq Dom (g) \times F\), so, hereafter, we frequently identify an element of \(S\) with the corresponding map, which means that we implicitly go between \(S'\) and \(S\): strict speaking, we should define the bijection between \(S'\) and \(S\) and explicitly go between \(S'\) and \(S\), but that would rather complicate the notations.
Let \(S\) have the ordering as containment.
\(S\) is a partially-ordered set, by the proposition that any set with the ordering as containment is a partially-ordered set.
\(T\) is normal, by the proposition that any paracompact Hausdorff topological space is normal.
\(S'\), so, \(S\), is not empty, because a \(g \in S'\) can be constructed like this.
Let \(l' \in L\) be any.
\(T = V_{l'} \cup \cup_{l \in L \setminus \{l'\}} V_l\), so, \(T \setminus (\cup_{l \in L \setminus \{l'\}} V_l) \subseteq V_{l'}\), because for each \(t \in T \setminus (\cup_{l \in L \setminus \{l'\}} V_l)\), \(t \notin \cup_{l \in L \setminus \{l'\}} V_l\), so, \(t \in V_{l'}\).
\(T \setminus (\cup_{l \in L \setminus \{l'\}} V_l) \subseteq T\) is closed and there is an open subset, \(U \subseteq T\), such that \(T \setminus (\cup_{l \in L \setminus \{l'\}} V_l) \subseteq U \subseteq \overline{U} \subseteq V_{l'}\), by 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.
There is a continuous map, \(f: T \to [0, 1]\), such that \(f (T \setminus (\cup_{l \in L \setminus \{l'\}} V_l)) = \{1\}\) and \(f (T \setminus U) = \{0\}\), by the proposition that for any normal topological space, any closed subset, and any open subset that contains the closed subset, there is a continuous map into any closed Interval that maps the closed subset to any boundary of the closed interval and the complement of the open subset to the other boundary (Urysohn's lemma).
So, \(f^{-1} ([0, 1] \setminus \{0\}) \subseteq U\), and \(\overline{f^{-1} ([0, 1] \setminus \{0\})} \subseteq \overline{U} \subseteq V_{l'}\).
As \(T \setminus (\cup_{l \in L \setminus \{l'\}} V_l) \subseteq f^{-1} ([0, 1] \setminus \{0\})\), \(T = f^{-1} ([0, 1] \setminus \{0\}) \cup (\cup_{l \in L \setminus \{l'\}} V_l)\), because for each \(t \in T\), when \(t \notin f^{-1} ([0, 1] \setminus \{0\})\), \(t \notin T \setminus (\cup_{l \in L \setminus \{l'\}} V_l)\), so, \(t \in \cup_{l \in L \setminus \{l'\}} V_l\).
So, \(g: \{l'\} \to F, l' \mapsto f \in S'\).
Step 2:
There is a maximal chain of \(S\), \(C \subseteq S\), by the Hausdorff maximal principle that any chain of any partially-ordered set is contained in a maximal chain: take any \(1\)-point subset as the initial chain.
Let us take \(s := \cup C\): each element of \(C\) is an element of \(S\), which is the relation of a map.
Step 3:
For each \((l, g (l)) \in s\), there is no other element of \(s\) with the same \(l\), because while \((l, g (l))\) is from \(g \in C\), for any other \(g' \in C\), \(g \subset g'\) or \(g' \subset g\), because \(C\) is a chain of \(S\), and while the domain of \(g'\) may or may not contain \(l\), if \(g'\) does, \(g' (l) = g (l)\), because otherwise, \(g \subset g'\) nor \(g' \subset g\) would be possible: as has been warned above, \(g\) or \(g'\) is conveniently regarded to be an element of \(S\) or to be the corresponding element of \(S'\): likewise expressions will be used below without any further warning.
So, \(s\) corresponds to a map, \(s: Dom (s) \to F\) for \(Dom (s) = \cup_{c \in C} Dom (c) \subseteq L\).
For each \(l \in Dom (s)\), \(\overline{s (l)^{-1} ([0, 1] \setminus \{0\})} \subseteq V_l\) holds, because \(l \in Dom (c)\) for a \(c \in C\) and \(s (l) = c (l)\).
Let us see that \(T = (\cup_{l \in Dom (s)} s (l)^{-1} ([0, 1] \setminus \{0\})) \cup (\cup_{l \in L \setminus Dom (s)} V_l)\).
Let us suppose otherwise.
There would a \(t \in T\) such that \(t \notin \cup_{l \in Dom (s)} s (l)^{-1} ([0, 1] \setminus \{0\})\) and \(t \notin \cup_{l \in L \setminus Dom (s)} V_l\).
There would be a nonempty finite \(\{l_1, ..., l_n\} \subseteq L\) such that \(t \in V_l\) for each \(l \in \{l_1, ..., l_n\}\) and \(t \notin V_l\) for each \(l \in L \setminus \{l_1, ..., l_n\}\), because \(\{V_l \vert l \in L\}\) is a locally finite open cover of \(T\).
For each \(l \in \{l_1, ..., l_n\}\), \(l \in Dom (s)\), because otherwise, \(l \in L \setminus Dom (s)\), so, \(t \in V_l \subseteq \cup_{l \in L \setminus Dom (s)} V_l\), a contradiction against \(t \notin \cup_{l \in L \setminus Dom (s)} V_l\).
So, \(l_1 \in Dom (c_1), ..., l_n \in Dom (c_n)\) for some \(c_1, ..., c_n \in C\), and as \(C\) is a chain, \(Dom (c_1) \subset ... \subset Dom (c_n)\) without loss of generality (if not, reorder \(\{l_1, ..., l_n\}\), by the proposition that for any linearly-ordered set, any finite subset can be ordered in a line: \(C\) is a linearly-ordered set and containment of its elements equals containment of the domains of the elements regarded as maps). So, \(l_1, ..., l_n \in Dom (c_n)\) without loss of generality.
As \(T = (\cup_{l \in Dom (c_n)} c_n (l)^{-1} ([0, 1] \setminus \{0\})) \cup (\cup_{l \in L \setminus Dom (c_n)} V_l)\), \(t \in (\cup_{l \in Dom (c_n)} c_n (l)^{-1} ([0, 1] \setminus \{0\})) \cup (\cup_{l \in L \setminus Dom (c_n)} V_l)\), but for each \(l \in L \setminus Dom (c_n)\), \(l \notin \{l_1, ..., l_n\}\), so, \(t \notin V_l\), so, \(t \notin \cup_{l \in L \setminus Dom (c_n)} V_l\), so, \(t \in \cup_{l \in Dom (c_n)} c_n (l)^{-1} ([0, 1] \setminus \{0\})\), which would mean that \(t \in \cup_{l \in Dom (s)} s (l)^{-1} ([0, 1] \setminus \{0\})\), a contradiction against \(t \notin \cup_{l \in Dom (s)} s (l)^{-1} ([0, 1] \setminus \{0\})\).
So, \(T = (\cup_{l \in Dom (s)} s (l)^{-1} ([0, 1] \setminus \{0\})) \cup (\cup_{l \in L \setminus Dom (s)} V_l)\).
So, \(s \in S\).
Then, \(s \in C\), because otherwise, \(C \subset C \cup \{s\}\) while \(C \cup \{s\}\) would be a chain, because for each \(c \in C\), \(c \subseteq \cup C = s\), which would imply \(c \subset s\), because \(c \neq s\), because \(s \notin C\), a contradiction against that \(C\) was maximal.
Let us see that \(Dom (s) = L\).
Let us suppose otherwise.
Let \(l' \in L \setminus Dom (s)\).
\(T = (\cup_{l \in Dom (s)} s (l)^{-1} ([0, 1] \setminus \{0\})) \cup (\cup_{l \in L \setminus Dom (s)} V_l) = (\cup_{l \in Dom (s)} s (l)^{-1} ([0, 1] \setminus \{0\})) \cup (\cup_{l \in L \setminus Dom (s) \setminus \{l'\}} V_l) \cup V_{l'}\), and let \(W := (\cup_{l \in Dom (s)} s (l)^{-1} ([0, 1] \setminus \{0\})) \cup (\cup_{l \in L \setminus Dom (s) \setminus \{l'\}} V_l)\).
\(T \setminus W \subseteq T\) would be closed, and \(T \setminus W \subseteq V_{l'}\), because for each \(t \in T \setminus W\), \(t \notin W\), so, \(t \in V_{l'}\).
There would be an open \(X \subseteq T\) such that \(T \setminus W \subseteq X \subseteq \overline{X} \subseteq V_{l'}\), by 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.
There would be a continuous \(h: T \to [0, 1]\) such that \(h (T \setminus W) = \{1\}\) and \(h (T \setminus X) = \{0\}\), by the proposition that for any normal topological space, any closed subset, and any open subset that contains the closed subset, there is a continuous map into any closed Interval that maps the closed subset to any boundary of the closed interval and the complement of the open subset to the other boundary (Urysohn's lemma).
\(T \setminus W \subseteq h^{-1} ([0, 1] \setminus \{0\})\), so, \(W \cup h^{-1} ([0, 1] \setminus \{0\}) = T\), because for each \(t \in T\), \(t \in W\) or \(t \in T \setminus W\), but when \(t \in T \setminus W\), \(t \in h^{-1} ([0, 1] \setminus \{0\})\).
\(h^{-1} ([0, 1] \setminus \{0\}) \subseteq X\), because for each \(t \in h^{-1} ([0, 1] \setminus \{0\})\), \(h (t) \in [0, 1] \setminus \{0\}\), so, \(t \notin T \setminus X\), so, \(t \in X\). So, \(\overline{h^{-1} ([0, 1] \setminus \{0\})} \subseteq \overline{X} \subseteq V_{l'}\).
So, \(s \cup (l', h) \in S\).
\(C \cup \{s \cup (l', h)\}\) would be a chain, because for each \(c \in C\), \(c \subseteq s \subset s \cup (l', h)\), a contradiction against that \(C\) was maximal.
So, \(Dom (s) = L\).
Step 4:
Let us take \(\rho := \sum_{l \in L} s (l)\), which is valid, because \(s (l)^{-1} ([0, 1] \setminus \{0\}) \subseteq V_l\) and \(\{V_l \vert l \in L\}\) is locally finite.
For each \(t \in T\), \(\rho (t) \neq 0\), because \(T = \cup_{l \in L} s (l)^{-1} ([0, 1] \setminus \{0\})\), so, \(t \in s (l)^{-1} ([0, 1] \setminus \{0\})\) for an \(l \in L\), so, \(0 \lt s (l) (t)\).
Let for each \(l \in L\), let \(\rho_l := s (l) / \rho\), which is valid, because \(\rho (t) \neq 0\) for each \(t \in T\).
Step 5:
\(Supp (\rho_l) = Supp (s (l)) = \overline{s (l)^{-1} ([0, 1] \setminus \{0\})} \subseteq V_l\).
\(\rho_l\) is continuous, by the proposition that for any topological space and any set of continuous maps from the space into any Euclidean topological space such that the set of the preimages of nonzero is locally finite, the sum of the maps is continuous: \(\{Supp (\rho_l) \vert l \in L\}\) is locally finite, because \(\{V_l \vert l \in L\}\) is locally finite while \(Supp (\rho_l) \subseteq V_l\).
\(0 \le \rho_l \le 1\), because \(0 \lt \rho\) and \(0 \le s (l)\) and \(s (l) / \rho = s (l) / \sum_{l' \in L} s (l') \le 1\).
\(\sum_{l \in L} \rho_l = 1\), because \(\sum_{l \in L} \rho_l = \sum_{l \in L} s (l) / \rho = \rho / \rho = 1\).
So, \(\{\rho_l \vert l \in L\}\) is a partition of unity subordinate to \(\{V_l \vert l \in L\}\).
