Tough lemma on locally finite refinement

[Korybut]

TL;DR

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

 

[Korybut]

TL;DR Summary: Finite refinement of locally compact, Hausdorff and second countable topological space

Why there is such an integer k1? V1¯ is compact due to previous paragraphs but {Unk} is just a basis. Why closure cannot belong to infinite union of basic open sets?

It is clear now and it is very easy. Let closure of $V_1$ is covered by finite amount of open sets which are not necessary basic elements. Let some of this open sets are generated by infinite union of basic set. This is still a cover and due to compactness there is finite subcover of basic open sets.

Part that is not clear is weather $\lbrace U_n \rbrace$ from the original basis all have compact closure?

 

[Korybut]

Part that is not clear is weather {Un} from the original basis all have compact closure?

Not necessary. I can not the proof that was presented. I can pick any $U_n$ from the basis $\lbrace U_n \rbrace$ and choose any point say $y$ within $U_n$. Due to local compactness there are open $U$ and compact $C$ such that $x\in U \subseteq C$. However $U_n$ might not be a subset of this particular $U$.

After two days. Lemma is not tough afterall

 

[Korybut]

TL;DR Summary: Finite refinement of locally compact, Hausdorff and second countable topological space

Moreover, Wij∩Wlk=∅ if | j -k |> 4. Finally, ∪j=1∞∪i=1mjWij=S. Hence, the collection

I believe this should insist that $\lbrace W_i^j \rbrace$ is locally finite however I am confused. If $|j-k| <4$ then this intersection can be non empty so locally finite refinement is not presented. Or I miss something... Need help

I get why within 4 steps intersection is trivial
$\bar{V}_{j+1} \subseteq V_{j+2}\subseteq \bar{V}_{j+2}$
thus $\bar{V}_{j+1} \nsubseteq V_{j+5} \setminus \bar{V}_{j+2}$. And hence presented in the proof intersection in indeed trivial. Why others?

 

[jbergman]

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

To answer your first question, the answer is no. For example, we can use the intervals with rational endpoints as a base for $\mathbb R$, but also include the open set of the entire real line and it is still a countable set and the entire real line is not contained in any compact set.

 

[Korybut]

To answer your first question, the answer is no. For example, we can use the intervals with rational endpoints as a base for $\mathbb R$, but also include the open set of the entire real line and it is still a countable set and the entire real line is not contained in any compact set.

Thanks! Your example exhibits the obstruction I was missing.

How about the last step of the proof? I know it is technical, sorry for that

I believe this should insist that $\lbrace W_i^j \rbrace$ is locally finite however I am confused. If $|j-k| <4$ then this intersection can be non empty so locally finite refinement is not presented. Or I miss something... Need help

I get why within 4 steps intersection is trivial
$\bar{V}_{j+1} \subseteq V_{j+2}\subseteq \bar{V}_{j+2}$
thus $\bar{V}_{j+1} \nsubseteq V_{j+5} \setminus \bar{V}_{j+2}$. And hence presented in the proof intersection in indeed trivial. Why others?

 

[Korybut]

I believe this should insist that $\lbrace W_i^j \rbrace$ is locally finite however I am confused. If $|j-k| <4$ then this intersection can be non empty so locally finite refinement is not presented. Or I miss something... Need help

I get why within 4 steps intersection is trivial
$\bar{V}_{j+1} \subseteq V_{j+2}\subseteq \bar{V}_{j+2}$
thus $\bar{V}_{j+1} \nsubseteq V_{j+5} \setminus \bar{V}_{j+2}$. And hence presented in the proof intersection in indeed trivial. Why others?

Problem occur due to my terrible misunderstanding of what is locally finite space. In my notes I have "Topological space where each point is contained in finitely many open sets", don't know where I found this one. With correct definition everything is obviously fine with the proof.

 

[jbergman]

Problem occur due to my terrible misunderstanding of what is locally finite space. In my notes I have "Topological space where each point is contained in finitely many open sets", don't know where I found this one. With correct definition everything is obviously fine with the proof.

Looking back at your first post on this thread I don't see where it proves that the cover is locally finite. O only see that you can find a subcover consisting of open sets with compact closures.

It seems like there is still more work to show the existence of a cover that is locally finite.

 

[Korybut]

Looking back at your first post on this thread I don't see where it proves that the cover is locally finite. O only see that you can find a subcover consisting of open sets with compact closures.

It seems like there is still more work to show the existence of a cover that is locally finite.

Actually everything is finished or at least perfectly clear to me now. These sets $\lbrace W_i^j \rbrace$ are indeed locally finite refinement of arbitrary cover $\lbrace U_\alpha \rbrace_{\alpha\in A}$ (Not the best notation IMO since $\lbrace U_n \rbrace$ is basis for topology).