description/proof of that for topological space and open cover of space, if there are locally finite refinement of open cover and partition of unity subordinate to refinement, there is partition of unity subordinate to original cover
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of refinement of open cover of subset of topological space.
- The reader knows a definition of partition of unity subordinate to open cover of topological space.
- The reader admits the proposition that for any topological space and any set of maps from the space into any ring or module such that the set of the preimages of nonzero is locally finite, the support of the sum of the maps is contained in the union of the supports of the maps.
- The reader admits the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
- The reader admits the proposition that any map between topological spaces is continuous if the domain restriction of the map to each open set of a possibly uncountable open cover is continuous.
- The reader admits the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset.
- The reader admits the proposition that the preimages of any disjoint subsets under any map are disjoint.
Target Context
- The reader will have a description and a proof of 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.
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:
\(T\): \(= \{\text{ the topological spaces }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{U_j \vert j \in J\}\): \(\in \{\text{ the open coverts of } T\}\)
//
Statements:
\(\exists \{V_l \vert l \in L\} \in \{\text{ the locally finite refinements of } \{U_j \vert j \in J\}\}, \exists \{\mu_l \vert l \in L\} \in \{\text{ the partitions of unity subordinate to } \{V_l \vert l \in L\}\}\)
\(\implies\)
\(\exists \{\rho \vert j \in J\} \in \{\text{ the partitions of unity subordinate to } \{U_j \vert j \in J\}\}\)
//
2: Note
For \(\{\mu_l \vert l \in L\}\) to be a partition of unity subordinate to \(\{V_l \vert l \in L\}\), that \(\{Supp (\mu_l) \vert l \in L\}\) is locally finite does not need to be checked, because as \(Supp (\mu_l) \subseteq V_l\), \(\{Supp (\mu_l) \vert l \in L\}\) is inevitably locally finite as \(\{V_l \vert l \in L\}\) is locally finite: for each \(t \in T\), there is a neighborhood of \(t\), \(N_t\), such that \(N_t\) intersects only some finite \(V_l\) s, then, \(N_t\) can intersect only the corresponding \(Supp (\mu_l)\) s, because if \(N_t \cap V_l = \emptyset\), \(N_t \cap Supp (\mu_l) = \emptyset\).
According to some people, when 'existence of a partition of unity subordinate to an open cover' is claimed, that is in the meaning of that there are a locally finite refinement of the open cover and a partition of unity subordinate to the refinement, but when the claim holds, there is indeed a partition of unity subordinate to the cover by our definition of partition of unity subordinate to open cover of topological space, by this proposition.
3: Proof
Whole Strategy: Step 1: take any map, \(f: L \to J\), such that \(V_l \subseteq U_{f (l)}\); Step 2: take \(\rho_j: T \to \mathbb{R} = \sum_{l \in f^{-1} (\{j\})} \mu_l\); Step 3: see that \(\rho_j\) is continuous; Step 4: see that \(0 \le \rho_j \le 1\); Step 5: see that \(Supp (\rho_j) \subseteq U_j\); Step 6: see that \(\{Supp (\rho_j) \vert j \in J\}\) is locally finite; Step 7: see that \(\sum_{j \in J} \rho_j = 1\).
Step 1:
For each \(l \in L\), \(V_l \subseteq U_j\) for a \(j \in J\), so, there is a map, \(f': L \to Pow (J), l \mapsto \{j \in J \vert V_l \subseteq U_j\}\), where each \(f' (l)\) is nonempty.
By the axiom of choice, there is a map, \(f: L \to J\), such that for each \(l \in L\), \(f (l) \in f' (l)\), which means that \(V_l \subseteq U_{f (l)}\).
Step 2:
For each \(j \in J\), let us take \(\rho_j: T \to \mathbb{R} = \sum_{l \in f^{-1} (\{j\})} \mu_l\).
That is valid, because \(\{Supp (\mu_l) \vert l \in L\}\) is locally finite and \(\{Supp (\mu_l) \vert l \in f^{-1} (\{j\})\}\) is locally finite, even more: see Note for the proposition that for any topological space and any set of maps from the space into any ring or module such that the set of the preimages of nonzero is locally finite, the support of the sum of the maps is contained in the union of the supports of the maps.
Step 3:
Let us see that for each \(j \in J\), \(\rho_j\) is continuous.
For each \(t \in T\), there is a neighborhood of \(t\), \(N_t \subseteq T\), such that there is a finite \(L^`_t \subseteq L\) such that for each \(l \in L \setminus L^`_t\), \(N_t \cap Supp (\mu_l) = \emptyset\), then, there is an open neighborhood of \(t\), \(U_t \subseteq T\), such that \(U_t \subseteq N_t\), and \(U_t \cap Supp (\mu_l) = \emptyset\) for each \(l \in L \setminus L^`_t\).
\(\rho_j \vert_{U_t} = \sum_{l \in f^{-1} (\{j\}) \cap L^`_t} \mu_l \vert_{U_t}\), because for each \(l \in f^{-1} (\{j\})\) such that \(l \notin L^`_t\), \(\mu_l \vert_{U_t} = 0\), because \(U_t \cap Supp (\mu_l) = \emptyset\), which means that for each \(t' \in U_t\), \(t' \notin Supp (\mu_l)\), so, \(t' \notin {\mu_l}^{-1} (\mathbb{R} \setminus \{0\})\), which means that \(\mu_l (t') = 0\).
So, \(\rho_j \vert_{U_t}\) is continuous, as a finite sum of some continuous maps into \(\mathbb{R}\): each \(\mu_l \vert_{U_t}\) is continuous, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
\(\{U_t \vert t \in T\}\) is an open cover of \(T\).
So, \(\rho_j\) is continuous, by the proposition that any map between topological spaces is continuous if the domain restriction of the map to each open set of a possibly uncountable open cover is continuous.
Step 4:
\(0 \le \rho_j\), because \(0 \le \mu_l\) for each \(l \in f^{-1} (\{j\})\).
\(\rho_j \le 1\), because \(\sum_{l \in f^{-1} (\{j\})} \mu_l \le \sum_{l \in L} \mu_l = 1\).
Step 5:
Let us see that for each \(j \in J\), \(Supp (\rho_j) \subseteq U_j\).
\(Supp (\rho_j) \subseteq \cup_{l \in f^{-1} (\{j\})} Supp (\mu_l)\), by the proposition that for any topological space and any set of maps from the space into any ring or module such that the set of the preimages of nonzero is locally finite, the support of the sum of the maps is contained in the union of the supports of the maps.
\(\subseteq \cup_{l \in f^{-1} (\{j\})} V_l \subseteq \cup_{l \in f^{-1} (\{j\})} U_j = U_j\).
So, \(Supp (\rho_j) \subseteq U_j\).
Step 6:
Let us see that \(\{Supp (\rho_j) \vert j \in J\}\) is locally finite.
Let \(t \in T\) be any.
There is a neighborhood of \(t\), \(N_t \subseteq T\), such that there is a finite \(L^`_t \subseteq L\) such that for each \(l \in L \setminus L^`_t\), \(N_t \cap Supp (\mu_l) = \emptyset\).
\(f (L^`_t) \subseteq J\) is finite.
For each \(j \in J \setminus f (L^`_t)\), for each \(l \in f^{-1} (\{j\})\), \(l \notin L^`_t\), because if \(l \in L^`_t\), \(j = f (l) \in f (L^`_t)\), a contradiction.
So, while \(Supp (\rho_j) \subseteq \cup_{l \in f^{-1} (\{j\})} Supp (\mu_l)\) as has been seen above, for each \(j \in J \setminus f (L^`_t)\), \(N_t \cap Supp (\rho_j) \subseteq N_t \cap \cup_{l \in f^{-1} (\{j\})} Supp (\mu_l) = \cup_{l \in f^{-1} (\{j\})} (N_t \cap Supp (\mu_l))\), by the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset, \(\cup_{l \in f^{-1} (\{j\})} \emptyset\), because \(l \in L \setminus L^`_t\), \(= \emptyset\).
So, \(N_t \cap Supp (\rho_j) = \emptyset\) for each \(j \in J \setminus f (L^`_t)\).
So, \(\{Supp (\rho_j) \vert j \in J\}\) is locally finite.
Step 7:
Let us see that \(\sum_{j \in J} \rho_j = 1\).
\(\sum_{j \in J} \rho_j = \sum_{j \in J} \sum_{l \in f^{-1} (\{j\})} \mu_l\).
\(= \sum_{l \in L} \mu_l\), because in \(\sum_{j \in J} \sum_{l \in f^{-1} (\{j\})}\), each \(l \in L\) appears only once, because \(l \in f^{-1} (\{f (l)\})\) and \(l \notin f^{-1} (\{j\})\) for each \(j \neq f (l)\), because \(f^{-1} (\{f (l)\}) \cap f^{-1} (\{j\}) = \emptyset\), by the proposition that the preimages of any disjoint subsets under any map are disjoint.
\(= 1\).
