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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
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.
\(= 1\).
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for topological space and set of maps from space into ring or module s.t. set of preimages of nonzero is locally finite, support of sum of maps is contained in union of supports of maps
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of 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.
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:
\(T\): \(= \{\text{ the topological spaces }\}\)
\(S\): \(\in \{\text{ the rings }\} \cup \{\text{ the modules }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{f_j: T \to S \vert j \in J\}\): such that \(\{{f_j}^{-1} (S \setminus \{0\}) \vert j \in J\}\) is locally finite
\(\sum_{j \in J} f_j\): \(: T \to S, t \mapsto \sum_{j \in J^`_t} f_j (t) \text{ for any } J^`_t \in \{\text{ the finite subsets of } J\} \text{ such that } \forall j \in J \setminus J^`_t (f_j (t) = 0)\)
//
Statements:
\(Supp (\sum_{j \in J} f_j) \subseteq \cup_{j \in J} Supp (f_j)\)
//
2: Note
That \(\{{f_j}^{-1} (S \setminus \{0\}) \vert j \in J\}\) is locally finite implies that for each \(t \in T\), there is a finite \(J^`_t \subseteq J\) such that for each \(j \in J \setminus J^`_t\), \(f_j (t) = 0\), because there is a neighborhood of \(t\), \(N_t\), such that there is a finite \(J^`_t \subseteq J\) such that for each \(j \in J \setminus J^`_t\), \(N_t \cap {f_j}^{-1} (S \setminus \{0\}) = \emptyset\), then, \(f_j (t) = 0\), because otherwise, \(t \in N_t \cap {f_j}^{-1} (S \setminus \{0\})\), a contradiction.
And while there can be some multiple such \(J^`_t\) s, \(\sum_{j \in J^`_t} f_j (t)\) does not depend on the choice of \(J^`_t\), because any \(J^`_t\) needs to contain all the \(j\) s such that \(f_j (t) \neq 0\) and adding any other \(j\) s does not influence the result: it is natural to take the \(J^`\) such that for each \(j \in J^`\), \(f_j (t) \neq 0\).
So, the definition of \(\sum_{j \in J} f_j\) is valid.
The condition that \(\{Supp (f_j) \vert j \in J\}\) is locally finite is a sufficient condition for that \(\{{f_j}^{-1} (S \setminus \{0\}) \vert j \in J\}\) is locally finite, because as \({f_j}^{-1} (S \setminus \{0\}) \subseteq Supp (f_j)\), if \(N_t\) intersects only some finite \(Supp (f_j)\) s, \(N_t\) can intersect only the corresponding finite \({f_j}^{-1} (S \setminus \{0\})\) s: if \(N_t\) does not intersect a \(Supp (f_j)\), \(N_t\) does not intersect the corresponding \({f_j}^{-1} (S \setminus \{0\})\).
3: Proof
Whole Strategy: Step 1: see that \((\sum_{j \in J} f_j)^{-1} (S \setminus \{0\}) \subseteq \cup_{j \in J} {f_j}^{-1} (S \setminus \{0\})\); Step 2: see that \(\overline{\cup_{j \in J} {f_j}^{-1} (S \setminus \{0\})} = \cup_{j \in J} \overline{{f_j}^{-1} (S \setminus \{0\})}\).
Step 1:
\(Supp (\sum_{j \in J} f_j) = \overline{(\sum_{j \in J} f_j)^{-1} (S \setminus \{0\})}\), by the definition of support.
\(Supp (f_j) = \overline{{f_j}^{-1} (S \setminus \{0\})}\), by the definition of support.
Let us see that \((\sum_{j \in J} f_j)^{-1} (S \setminus \{0\}) \subseteq \cup_{j \in J} {f_j}^{-1} (S \setminus \{0\})\).
For each \(t \in (\sum_{j \in J} f_j)^{-1} (S \setminus \{0\})\), \((\sum_{j \in J} f_j) (t) \in S \setminus \{0\}\), which implies that \(f_j (t) \in S \setminus \{0\}\) for a \(j \in J\), so, \(t \in {f_j}^{-1} (S \setminus \{0\})\), so, \(t \in \cup_{j \in J} {f_j}^{-1} (S \setminus \{0\})\).
Step 2:
So, \(Supp (\sum_{j \in J} f_j) = \overline{(\sum_{j \in J} f_j)^{-1} (S \setminus \{0\})} \subseteq \overline{\cup_{j \in J} {f_j}^{-1} (S \setminus \{0\})} = \cup_{j \in J} \overline{{f_j}^{-1} (S \setminus \{0\})}\), 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, \(= \cup_{j \in J} Supp (f_j)\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that paracompact Hausdorff topological space is normal
Topics
About:
topological space
The table of contents of this article
Starting Context
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:
\(T\): \(= \{\text{ the paracompact Hausdorff topological spaces }\}\)
//
Statements:
\(T \in \{\text{ the normal topological spaces }\}\)
//
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.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that closed subspace of paracompact topological space is paracompact
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that any closed subspace of any paracompact topological space is paracompact.
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:
\(T'\): \(= \{\text{ the paracompact topological spaces }\}\)
\(T\): \(\in \{\text{ the closed subspaces of } T'\}\)
//
Statements:
\(T \in \{\text{ the paracompact topological spaces }\}\)
//
2: Proof
Whole Strategy: Step 1: for each open cover of \(T\), \(\{U_j \vert j \in J\}\), take the open cover of \(T'\), \(\{U'_J \vert j \in J\} \cup \{T' \setminus T\}\), where \(U_j = U'_j \cap T\); Step 2: take any locally finite refinement of \(\{U'_J \vert j \in J\} \cup \{T' \setminus T\}\), \(\{V'_l \vert l \in L\}\), and see that \(\{V'_l \cap T \vert l \in L\} \setminus \{\emptyset\}\) is a locally finite refinement of \(\{U_j \vert j \in J\}\).
Step 1:
Let \(\{U_j \vert j \in J\}\) where \(J\) is any possibly uncountable index set be any open cover of \(T\).
For each \(j \in J\), \(U_j = U'_j \cap T\) where \(U'_j \subseteq T'\) is an open subset, by the definition of topological subspace.
Let us take \(\{U'_J \vert j \in J\} \cup \{T' \setminus T\}\), which is an open cover of \(T'\), because \(U'_J \subseteq T'\) is open and \(T' \setminus T \subseteq T'\) is open, and for each \(t' \in T'\), \(t' \in T\) or \(t' \in T' \setminus T\), and when \(t' \in T\), \(t' \in U_j\) for a \(j \in J\), so, \(t' \in U'_j \cap T \subseteq U'_j\), and when \(t' \in T' \setminus T\), \(t' \in T' \setminus T\).
Step 2:
As \(T'\) is paracompact, there is a locally finite refinement of \(\{U'_J \vert j \in J\} \cup \{T' \setminus T\}\), \(\{V'_l \vert l \in L\}\).
Let us take \(\{V'_l \cap T \vert l \in L\} \setminus \{\emptyset\} = \{V'_l \cap T \vert l \in L^`\}\) where \(L^` \subseteq L\): in fact, \(\emptyset\) does not need to be removed although it is useless.
It is an open cover of \(T\), because \(V'_l \cap T\) is open on \(T\), by the definition of topological subspace, and for each \(t \in T\), as \(t \in T'\), \(t \in V'_l\) for an \(l \in L\), and \(t \in V'_l \cap T\), while \(l \in L^`\), because \(V'_l \cap T \neq \emptyset\).
For each \(l \in L^`\), \(V'_l \subseteq U'_j\) for a \(j \in J\) (\(V'_l \subseteq T' \setminus T\) is impossible, because otherwise, \((V'_l \cap T) \subseteq ((T' \setminus T) \cap T) = \emptyset\)), so, \(V'_l \cap T \subseteq U'_j \cap T = U_j\), so, \(\{V'_l \cap T \vert l \in L^`\}\) is a refinement of \(\{U_j \vert j \in J\}\).
Let us see that \(\{V'_l \cap T \vert l \in L^`\}\) is locally finite on \(T\).
Let \(t \in T\) be any.
\(t \in T'\), so, there is a neighborhood of \(t\), \(N'_t \subseteq T'\), such that \(N'_t\) intersects only some finite \(\{V'_{l_1}, ..., V'_{l_n}\}\) among \(\{V'_l \vert l \in L\}\), because \(\{V'_l \vert l \in L\}\) is locally finite.
\(N'_t \cap T \subseteq T\) is a neighborhood of \(t\), by the proposition that for any topological space and any point on any subspace, the intersection of any neighborhood of the point on the base space and the subspace is a neighborhood on the subspace.
Then, \(N'_t \cap T\) can intersect only \(\{V'_{l_1} \cap T, ..., V'_{l_n} \cap T\}\) among \(\{V'_l \cap T \vert l \in L\}\), because when \(N'_t \cap V'_l = \emptyset\), \(N'_t \cap V'_l \cap T = \emptyset \cap T = \emptyset\), while \(N'_t \cap V'_l \cap T = N'_t \cap V'_l \cap T \cap T = (N'_t \cap T) \cap (V'_l \cap T)\), then, \(N'_t \cap T\) can intersect only \(\{V'_{l_1} \cap T, ..., V'_{l_n} \cap T\} \setminus \{\emptyset\}\) among \(\{V'_l \cap T \vert l \in L^`\}\).
So, \(\{V'_l \cap T \vert l \in L^`\}\) is locally finite on \(T\).
So, \(T\) is paracompact.
3: Note
\(T \subseteq T'\) needs to be closed, because otherwise, we could not construct an open cover of \(T'\) as \(\{U'_J \vert j \in J\} \cup \{T' \setminus T\}\), and we would need to construct like \(\{U'_J \vert j \in J\} \cup \{U'_m \vert m \in M\}\), but taking a locally finite refinement of it, \(\{V'_l \vert l \in L\}\), \(\{V'_l \cap T \vert l \in L\}\) would not be guaranteed to be a refinement of \(\{U_j \vert j \in J\}\), because it might be \(V'_l \cap T \subseteq V'_l \subseteq U'_m\) for an \(m \in M\), and \(V'_l \cap T \subseteq U'_m \cap T\) would not imply \(U'_m \cap T \in \{U_j \vert j \in J\}\).
References
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definition of paracompact topological space
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of paracompact topological space.
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:
\(*T\): \(\in \{\text{ the topological spaces }\}\)
//
Conditions:
\(\forall \{U_j \vert j \in J\} \in \{\text{ the open covers of } T\} (\exists \{V_l \vert l \in L\} \in \{\text{ the refinements of } \{U_j \vert j \in J\}\} (\{V_l \vert l \in L\} \in \{\text{ the locally finite sets of subsets of } T\}))\)
//
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
definition of refinement of open cover of subset of topological space
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of refinement of open cover of subset of topological space.
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:
\( T\): \(\in \{\text{ the topological spaces }\}\)
\( S\): \(\subseteq T\)
\( J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\( \{U_j \in \{\text{ the open subsets of } T\} \vert j \in J\}\): such that \(S \subseteq \cup_{j \in J} U_j\)
\( L\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(*\{V_l \in \{\text{ the open subsets of } T\} \vert l \in L\}\): such that \(S \subseteq \cup_{l \in L} V_l\)
//
Conditions:
\(\forall l \in L (\exists j \in J (V_l \subseteq U_j))\)
//
2: Note
Some people may require \(\{V_l \vert l \in L\}\) to be just a cover of \(S\) instead of being an open cover by "refinement" and then, they will say "open refinement" for this definition.
This definition requires \(\{V_l \vert l \in L\}\) to be an open cover, because we usually naturally expect an open cover for any open cover, \(\{U_j \vert j \in J\}\).
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that locally compact Hausdorff topological space is open subspace of its \(1\)-point compactification
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that any locally compact Hausdorff topological space is an open subspace of its \(1\)-point compactification.
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:
\(T\): \(\in \{\text{ the locally compact Hausdorff topological spaces }\}\)
\(T^+\): \(= \text{ the 1-point compactification }\)
//
Statements:
\(T \in \{\text{ the open subspaces of } T^+\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(T\) is the subspace of the Hausdorff \(T^+\); Step 2: see that \(T = T^+ \setminus \{\infty\}\) is open on \(T^+\).
Step 1:
\(T^+\) is a Hausdorff topological space and \(T\) is the topological subspace of \(T^+\), by the proposition that for any locally compact Hausdorff topological space, the topology of the \(1\)-point compactification is the only topology that makes the \(1\)-point-augmented set compact Hausdorff with the original space as the subspace.
Step 2:
\(\{\infty\} \subseteq T^+\) is closed, by the proposition that for any Hausdorff topological space, any 1 point subset is closed.
So, \(T = T^+ \setminus \{\infty\}\) is open on \(T^+\).
So, \(T\) is an open topological subspace of \(T^+\).
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for Hausdorff topological space, if there is locally compact Hausdorff topological space of which space is locally closed subspace, space is locally compact
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any Hausdorff topological space, if there is a locally compact Hausdorff topological space of which the space is a locally closed subspace, the space is locally compact.
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:
\(T\): \(\in \{\text{ the Hausdorff topological spaces }\}\)
//
Statements:
\(\exists T' \in \{\text{ the locally compact Hausdorff topological spaces }\} (T \in \{\text{ the locally closed subspaces of } T'\})\)
\(\implies\)
\(T \in \{\text{ the locally compact topological spaces }\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(T = C' \cap U'\) for a closed \(C' \subseteq T'\) and an open \(U' \subseteq T'\); Step 2: see that \(C'\) is locally compact and that \(T\) is a locally compact subspace of \(C'\).
Step 1:
There are a closed \(C' \subseteq T'\) and an open \(U' \subseteq T'\) such that \(T = C' \cap U'\), by the proposition that any topological subspace is locally closed if and only if it is the intersection of a closed subset and an open subset of the base space.
Step 2:
\(C' \subseteq T'\) is a locally compact Hausdorff topological subspace, by the proposition that any closed subspace of any locally compact topological space is locally compact and the proposition that any subspace of any Hausdorff topological space is Hausdorff.
\(T = C' \cap U' \subseteq C'\), as the topological subspace of \(C'\), is an open subspace, by the definition of topological subspace, and is a locally compact topological subspace of \(C'\), by the proposition that any open subspace of any locally compact Hausdorff topological space is locally compact.
But \(T\) as the topological subspace of \(C'\) is \(T\) as the topological subspace of \(T'\), by the proposition that in any nest of topological subspaces, the openness of any subset on any subspace does not depend on the superspace of which the subspace is regarded to be a subspace.
As \(T\) we are talking about is the topological subspace of \(T'\), \(T\) is locally compact.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
definition of locally closed topological subspace
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of locally closed topological subspace.
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:
\( T'\): \(\in \{\text{ the topological spaces }\}\)
\(*T\): \(\subseteq T'\), with the subspace topology of \(T'\)
//
Conditions:
\(\forall t \in T (\exists U'_t \in \{\text{ the open neighborhoods of } t \text{ on } T' \text{ with the subspace topology of } T'\} (U'_t \cap T \in \{\text{ the closed subsets of } U'_t\}))\)
//
2: Note
When \(T \subseteq T'\) is closed, \(T\) is locally closed, because for whatever open neighborhood of \(t\), \(U'_t \subseteq T'\), for example, \(U'_t = T'\), \(U'_t \cap T\) is closed on \(U'_t\), by the proposition that any subset on any topological subspace is closed if and only if there is a closed set on the base space whose intersection with the subspace is the subset.
When \(T \subseteq T'\) is open, \(T\) is locally closed, because there is an open neighborhood of \(t\), \(U'_t \subseteq T'\), such that \(U'_t \subseteq T\), by the local criterion for openness, and \(U'_t \cap T = U'_t \subseteq U'_t\) is closed.
Any interval, \(T = I \subseteq \mathbb{R} = T'\), where \(I = (r_1, r_2), (r_1, r_2], [r_1, r_2), \text{ or } [r_1, r_2]\) where \(r_1\) may be \(- \infty\) when \(I\) is lower open and \(r_2\) may be \(\infty\) when \(I\) is upper open, is locally closed, because while any open or closed case is already known to be locally closed, for \((r_1, r_2]\), when \(t \neq r_2\), there is a \(U'_t \subseteq (r_1, r_2]\), and \(U'_t \cap T = U'_t \subseteq U'_t\) is closed, and when \(t = r_2\), there is a \(U'_t = B_{r_2, \epsilon}\) such that \(r_1 \lt r_2 - \epsilon\), and \(U'_t \cap T = (r_2 - \epsilon, r_2]\), which is closed on \(B_{r_2, \epsilon}\), because \((r_2 - \epsilon, r_2] = B_{r_2, \epsilon} \cap [r_2 - \epsilon, r_2]\); for \([r_1, r_2)\), likewise.
Then, what are not locally closed?
For example, \(T = \mathbb{Q} \subseteq \mathbb{R} = T'\) is not locally closed, because for any \(q \in \mathbb{Q}\), for any \(U'_q \subseteq \mathbb{R}\), \(U'_q \cap \mathbb{Q} \subseteq U'_q\) is not closed, because \(U'_q \setminus (U'_q \cap \mathbb{Q}) \subseteq U'_q\) is not open, because for each \(r \in U'_q \setminus (U'_q \cap \mathbb{Q})\), any \(B'_{r, \epsilon} \subseteq \mathbb{R}\) such that \(B'_{r, \epsilon} \subseteq U'_q\) contains a rational number, so, \(B'_{r, \epsilon} \subseteq U'_q \setminus (U'_q \cap \mathbb{Q})\) does not hold.
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that proper continuous map between locally compact Hausdorff topological spaces is closed
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that any proper continuous map between any locally compact Hausdorff topological spaces is closed.
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:
\(T_1\): \(\in \{\text{ the locally compact Hausdorff topological spaces }\}\)
\(T_2\): \(\in \{\text{ the locally compact Hausdorff topological spaces }\}\)
\(f\): \(: T_1 \to T_2\), \(\in \{\text{ the proper maps }\} \cap \{\text{ the continuous maps }\}\)
//
Statements:
\(f \in \{\text{ the closed maps }\}\)
//
2: Proof
Whole Strategy: Step 1: take the \(1\)-point compactifications, \({T_1}^+\) and \({T_2}^+\), and the continuous extension of \(f\), \(f': {T_1}^+ \to {T_2}^+\); Step 2: see that for each closed \(C_1 \subseteq T_1\), \(f (C_1) = f' (C_1 \cup \{\infty\}) \cap T_2\) is closed on \(T_2\).
Step 1:
Let us take the \(1\)-point compactification of \(T_1\), \({T_1}^+\), and the \(1\)-point compactification of \(T_2\), \({T_2}^+\).
\({T_1}^+\) and \({T_2}^+\) are some compact Hausdorff topological spaces such that \(T_1\) and \(T_2\) are their topological subspaces, by the proposition that for any locally compact Hausdorff topological space, the topology of the \(1\)-point compactification is the only topology that makes the \(1\)-point-augmented set compact Hausdorff with the original space as the subspace.
Let us define \(f': {T_1}^+ \to {T_2}^+\) as the extension of \(f\) such that \(f' (\infty) = \infty\).
Let us see that \(f'\) is continuous.
Let \(U_2 \subseteq {T_2}^+\) be any open subset.
\(U_2\) is an open subset of \(T_2\) or \(U_2 = {T_2}^+ \setminus K\) where \(K \subseteq T_2\) is a compact subset.
When \(U_2\) is an open subset of \(T_2\), \(f'^{-1} (U_2) = f^{-1} (U_2)\), by the proposition that for any map and its any extension that maps the extended area outside the original codomain, the extension preimage of any subset of the extension codomain is the union of the original map preimage of the intersection of the subset and the original codomain and the extension preimage of the subset minus the original codomain, which is open on \(T_1\), because \(f\) is continuous, so, is open on \({T_1}^+\).
When \(U_2 = {T_2}^+ \setminus K\), \(f'^{-1} (U_2) = f'^{-1} ({T_2}^+ \setminus K) = f'^{-1} ({T_2}^+) \setminus f'^{-1} (K)\), by the proposition that for any map, the preimage of any subset minus any subset is the preimage of the 1st subset minus the preimage of the 2nd subset, \(= {T_1}^+ \setminus f^{-1} (K)\), by the proposition that the preimage of the whole codomain of any map is the whole domain and the proposition that for any map and its any extension that maps the extended area outside the original codomain, the extension preimage of any subset of the extension codomain is the union of the original map preimage of the intersection of the subset and the original codomain and the extension preimage of the subset minus the original codomain, but \(f^{-1} (K) \subseteq T_1\) is compact, because \(f\) is proper, so, \({T_1}^+ \setminus f^{-1} (K)\) is open on \({T_1}^+\).
So, \(f'\) is continuous.
Step 2:
Let \(C_1 \subseteq T_1\) be any closed subset.
\(f (C_1) = f' (C_1 \cup \{\infty\}) \cap T_2\), by the proposition that for any map and its any extension that maps the extended area outside the original codomain, the original map image of any subset of the original domain is the intersection of the extension image of the union of the subset and the extended area and the original codomain.
But \(C_1 \cup \{\infty\} \subseteq {T_1}^+\) is closed, because \({T_1}^+ \setminus (C_1 \cup \{\infty\}) = ({T_1}^+ \setminus C_1) \cap ({T_1}^+ \setminus \{\infty\})\), by the proposition that for any set, any subset minus the union of any subsets is the intersection of the 1st subset minus the 2nd chunk of subsets, \(= ({T_1}^+ \setminus C_1) \cap T_1 = ({T_1}^+ \cap T_1) \setminus (C_1 \cap T_1)\), by the proposition that the intersection of any set minus any set and any set is the intersection of the 1st set and the 3rd set minus the intersection of the 2nd set and the 3rd set, \(= T_1 \setminus C_1\), which is open on \(T_1\), so, is open on \({T_1}^+\).
So, \(C_1 \cup \{\infty\}\) is compact on \({T_1}^+\), by the proposition that any closed subset of any compact topological space is compact.
\(f' (C_1 \cup \{\infty\}) \subseteq {T_2}^+\) is compact, by the proposition that for any continuous map between any topological spaces, the image of any compact subset of the domain is a compact subset of the codomain, so, is closed on \({T_2}^+\), by the proposition that any compact subset of any Hausdorff topological space is closed.
So, \(f (C_1) = f' (C_1 \cup \{\infty\}) \cap T_2\) is closed on \(T_2\), by the proposition that any subset on any topological subspace is closed if and only if there is a closed set on the base space whose intersection with the subspace is the subset.
So, \(f\) is closed.
References
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