- #1
Korybut
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- TL;DR Summary
- Finite refinement of locally compact, Hausdorff and second countable topological space
Hello!
I have some troubles diving in the proof of this lemma
Lemma. Let ##S## be locally compact, Hausdorff and second countable. Then every open cover ##\lbrace U_\alpha \rbrace## of ##S## has a countable, locally finite refinement consisting of open sets with compact closures.
Proof. Since ##S## is second countable, there exists a countable family ##\lbrace U_n \rbrace## of open sets in ##S##, which forms a basis for the topology of ##S##. Let ##\lbrace U_{n_k} \rbrace## be a subcollection consisting of sets with compact closures.
The assumption that ##S## is Hausdorff and locally compact implies that ##\lbrace U_{n_k} \rbrace ## is a basis for the topology of ##S##. This can be seen as follows. Since ##S## is locally compact, given a point ##x \in S## there exists a compact set ##C## in ##S## containing an open neighbourhood ##U## of ##x##. That is, ##x\in U \subseteq C##. Since ##C## is a compact subset of the Hausdorff space ##S##, it is closed. Hence, ##\bar{U} \subseteq C##, which implies that ##\bar{U}## is compact. The assumption that ##\lbrace U_n ## is a basis implies that there exists ##n_x## such that ##x\in U_{n_x} \subseteq U##. Hence, ##U_{n_x}## is compact. Therefore, ##U_{n_x}\in \lbrace U_{n_k} \rbrace##.
These two paragraphs are almost clear. The only question I have is the following. Since point ##x## is not fixed I can change it and pick corresponding ##U_{n_x}##. Does it imply that every ## U_n## from the collection that form a basis has compact closure? I did not manage to find counter example
We set ##V_{-1} = V_0 = \emptyset## and take ##V_1 = U_{n_1}##. There exists a smallest integer ##k_1## such that ##\bar{V}_1\subseteq U_{n_1}\cup ... \cup U_{n_{k_1}}##. We now set ##V_2 = V_1 \cup U_{n_2}\cup ... \cup U_{n_{k_1}}##. Continuing in this way, we obtain a sequence of open sets ##V_j = U_{n_1}\cup ... \cup U_{n_{k_j}}## for every ##j\in\mathbb{N}##, where ##n_{k_j}## is the smallest integer such that ##\bar{V}_{j-1}\subseteq V_{j-1} \cup ... U_{n_{k_{j-1}+1 }}\cup ...\cup U_{n_{k_j}}## . For each ##j##, the closure ##\bar{V_j}## of ##V_j## is contained in ##V_{j+1}## and ##\cup_{j=1}^\infty V_j=S##. For each ##j\in \mathbb{N}##, the set ##\bar{V_j}\setminus V_{j-1}## is compact and is contained in the open set ##V_{j+1}\setminus\bar{V}_{j-2}##.
Why there is such an integer ##k_1##? ##\bar{V_1}## is compact due to previous paragraphs but ##\lbrace U_{n_k} \rbrace## is just a basis. Why closure cannot belong to infinite union of basic open sets?
Let ##\lbrace U_\alpha\rbrace_{\alpha \in A}## be an arbitrary open cover of ##S##. Hence, ##\lbrace U_\alpha \cap (V_{j+1}\setminus\bar{V}_{j-2})\rbrace_{\alpha \in A}## is an open cover of the compact set ##\bar{V}_j\setminus V_{j-1}## and it admits a finite subcover. We denote by ##\lbrace W_1^j, . . . , W_{m_j}^j\rbrace## the finite collection of sets in ##\lbrace U_\alpha \cap (V_{j+1}\setminus\bar{V}_{j-2})\rbrace_{\alpha \in A}## which cover ##\bar{V}_j\setminus V_{j-1}##. Each ##W_i^j## is contained in a compact set ##\bar{V}_{j+1}##. Hence, ##\bar{W}_i^j## is compact. Moreover, for some ##\alpha \in A##, ##W_i^j =U_\alpha \cap (V_{j+1}\setminus\bar{V}_{j-2})##, so that ##W_i^j\subseteq U_\alpha##. Moreover, ##W_i^j \cap W_l^k = \emptyset## if | j -k |> 4. Finally, ##\cup_{j=1}^\infty \cup_{i=1}^{m_j} W_i^j = S##. Hence, the collection ##\lbrace W_i^j \vert i = 1, . . . , m_j, j\in \mathbb{N}\rbrace## is a countable, locally finite refinement of ##\lbrace U_\alpha \rbrace## and consists of open sets with compact closures.
I am not here yet to ask any reasonable questions. But looks fine after couple of readings.
Many thanks in advance
I have some troubles diving in the proof of this lemma
Lemma. Let ##S## be locally compact, Hausdorff and second countable. Then every open cover ##\lbrace U_\alpha \rbrace## of ##S## has a countable, locally finite refinement consisting of open sets with compact closures.
Proof. Since ##S## is second countable, there exists a countable family ##\lbrace U_n \rbrace## of open sets in ##S##, which forms a basis for the topology of ##S##. Let ##\lbrace U_{n_k} \rbrace## be a subcollection consisting of sets with compact closures.
The assumption that ##S## is Hausdorff and locally compact implies that ##\lbrace U_{n_k} \rbrace ## is a basis for the topology of ##S##. This can be seen as follows. Since ##S## is locally compact, given a point ##x \in S## there exists a compact set ##C## in ##S## containing an open neighbourhood ##U## of ##x##. That is, ##x\in U \subseteq C##. Since ##C## is a compact subset of the Hausdorff space ##S##, it is closed. Hence, ##\bar{U} \subseteq C##, which implies that ##\bar{U}## is compact. The assumption that ##\lbrace U_n ## is a basis implies that there exists ##n_x## such that ##x\in U_{n_x} \subseteq U##. Hence, ##U_{n_x}## is compact. Therefore, ##U_{n_x}\in \lbrace U_{n_k} \rbrace##.
These two paragraphs are almost clear. The only question I have is the following. Since point ##x## is not fixed I can change it and pick corresponding ##U_{n_x}##. Does it imply that every ## U_n## from the collection that form a basis has compact closure? I did not manage to find counter example
We set ##V_{-1} = V_0 = \emptyset## and take ##V_1 = U_{n_1}##. There exists a smallest integer ##k_1## such that ##\bar{V}_1\subseteq U_{n_1}\cup ... \cup U_{n_{k_1}}##. We now set ##V_2 = V_1 \cup U_{n_2}\cup ... \cup U_{n_{k_1}}##. Continuing in this way, we obtain a sequence of open sets ##V_j = U_{n_1}\cup ... \cup U_{n_{k_j}}## for every ##j\in\mathbb{N}##, where ##n_{k_j}## is the smallest integer such that ##\bar{V}_{j-1}\subseteq V_{j-1} \cup ... U_{n_{k_{j-1}+1 }}\cup ...\cup U_{n_{k_j}}## . For each ##j##, the closure ##\bar{V_j}## of ##V_j## is contained in ##V_{j+1}## and ##\cup_{j=1}^\infty V_j=S##. For each ##j\in \mathbb{N}##, the set ##\bar{V_j}\setminus V_{j-1}## is compact and is contained in the open set ##V_{j+1}\setminus\bar{V}_{j-2}##.
Why there is such an integer ##k_1##? ##\bar{V_1}## is compact due to previous paragraphs but ##\lbrace U_{n_k} \rbrace## is just a basis. Why closure cannot belong to infinite union of basic open sets?
Let ##\lbrace U_\alpha\rbrace_{\alpha \in A}## be an arbitrary open cover of ##S##. Hence, ##\lbrace U_\alpha \cap (V_{j+1}\setminus\bar{V}_{j-2})\rbrace_{\alpha \in A}## is an open cover of the compact set ##\bar{V}_j\setminus V_{j-1}## and it admits a finite subcover. We denote by ##\lbrace W_1^j, . . . , W_{m_j}^j\rbrace## the finite collection of sets in ##\lbrace U_\alpha \cap (V_{j+1}\setminus\bar{V}_{j-2})\rbrace_{\alpha \in A}## which cover ##\bar{V}_j\setminus V_{j-1}##. Each ##W_i^j## is contained in a compact set ##\bar{V}_{j+1}##. Hence, ##\bar{W}_i^j## is compact. Moreover, for some ##\alpha \in A##, ##W_i^j =U_\alpha \cap (V_{j+1}\setminus\bar{V}_{j-2})##, so that ##W_i^j\subseteq U_\alpha##. Moreover, ##W_i^j \cap W_l^k = \emptyset## if | j -k |> 4. Finally, ##\cup_{j=1}^\infty \cup_{i=1}^{m_j} W_i^j = S##. Hence, the collection ##\lbrace W_i^j \vert i = 1, . . . , m_j, j\in \mathbb{N}\rbrace## is a countable, locally finite refinement of ##\lbrace U_\alpha \rbrace## and consists of open sets with compact closures.
I am not here yet to ask any reasonable questions. But looks fine after couple of readings.
Many thanks in advance
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