323: Locally Compact Hausdorff Topological Space Is Paracompact iff Space Is Disjoint Union of Open \sigma-Compact Subspaces

<The previous article in this series | The table of contents of this series | The next article in this series>

A description/proof of that locally compact Hausdorff topological space is paracompact iff space is disjoint union of open $\sigma$-compact subspaces

---

Topics

About: topological space

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

---

Starting Context

• The reader knows a definition of locally compact topological space.
• The reader knows a definition of Hausdorff topological space.
• The reader knows a definition of paracompact topological space.
• The reader knows a definition of $\sigma$-compact topological space.
• The reader admits the proposition that any subspace of any Hausdorff topological space is Hausdorff.
• The reader admits the proposition that any open subspace of any locally compact Hausdorff topological space is locally compact.
• The reader admits the proposition that the topological sum of any possibly uncountable number of paracompact topological spaces is paracompact.
• The reader admits the proposition that for any topological space, the union of any finite compact subsets is compact.
• The reader admits the proposition that any closed subset of any compact topological space is compact.
• The reader admits the proposition that any compact subset of any Hausdorff topological space is closed.
• The reader admits the proposition that for any topological space, any compact subset of any subspace is compact on the base space.
• The reader admits the proposition that any compact Hausdorff topological space is normal.
• The reader admits the proposition that any open set minus any closed set is open.
• The reader admits the proposition that for any locally finite cover of any topological space, any compact subset intersects only finite elements of the cover.
• The reader admits the proposition that the closure of the union of any finite number of subsets is the union of the closures of the subsets.
• The reader admits the proposition that for any locally finite open cover of any topological space, the closure of the union of any possibly uncountable open sets in the cover is the union of the closures of the open sets.
• The reader admits the proposition that for any disjoint subset and open set, the closure of the subset and the open set are disjoint.
• The reader admits the proposition that for any locally compact Hausdorff topological space, around any point, there is an open neighborhood whose closure is compact.

---

Target Context

• The reader will have a description and a proof of the proposition that any locally compact Hausdorff topological space is paracompact if and only if the space is the union of some disjoint open $\sigma$-compact subspaces.

---

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: Description

For any locally compact Hausdorff topological space, $T$, $T$ is paracompact if and only if $T$ is the union of some disjoint open $\sigma$-compact subspaces, $\{T_{\alpha }|\alpha \in A\}$, where A is any possibly uncountable indices set.

---

2: Proof

Let us suppose that $T$ is the union of some disjoint open $\sigma$-compact subspaces, $\{T_{\alpha }|\alpha \in A\}$.

By the proposition that any subspace of any Hausdorff topological space is Hausdorff and the proposition that any open subspace of any locally compact Hausdorff topological space is locally compact, each $T_{\alpha }$ is locally compact Hausdorff. If each $T_{\alpha }$ is paracompact, $T=\{T_{\alpha }|\alpha \in A\}$ will be paracompact, by Note in the proposition that the topological sum of any possibly uncountable number of paracompact topological spaces is paracompact.

By the definition of $\sigma$-compactness, there is a countable number of compact subsets, $\{S_i|i\in J\}$ where $J$ is a countable indices set, such that $T_{\alpha }={\cup }_{i\in J}{S}_i$. Let us inductively define ${S_i}^{\prime }$ as ${S_0}^{\prime }:=S_0$ and for ${S_i}^{\prime }$ for $1\le i$, supposing that we have already defined a compact ${S_{i-1}}^{\prime }$, let us take an open cover of ${S_{i-1}}^{\prime }$ by taking around each point, $p\in {S_{i-1}}^{\prime }$, an open neighborhood whose closure is compact, which is possible by the proposition that for any locally compact Hausdorff topological space, around any point, there is an open neighborhood whose closure is compact, then, take a finite subcover, which is possible because ${S_{i-1}}^{\prime }$ is compact, and denoting the union of the closures of the open neighborhoods in the subcover as ${S_{i-1}}^{″}$, ${S_i}^{\prime }:=S_i\cup {S_{i-1}}^{″}$, which is compact on $T_{\alpha }$ by the proposition that for any topological space, the union of any finite compact subsets is compact. ${S_i}^{\prime }$ is closed by the proposition that any compact subset of any Hausdorff topological space is closed. If ${S_i}^{\prime }$ happens to equal ${S_{i-1}}^{\prime }$, let us omit it, so, we suppose ${S_{i-1}}^{\prime }\subset {S_i}^{\prime }$.

$\{{S_i}^{\prime }|i\in J\}$ covers $T_{\alpha }$ because every $S_i$ is contained in ${S_i}^{\prime }$ while $T_{\alpha }={\cup }_{i\in J}{S}_i$. ${S_i}^{\prime }\subseteq int{S_{i+1}}^{\prime }$, because ${S_i}^{\prime }$ is contained in its open cover, which is contained in ${S_{i+1}}^{\prime }$, but the open cover is open and is contained in the largest open set contained in ${S_{i+1}}^{\prime }$, which is $int{S_{i+1}}^{\prime }$.

Let us define ${S_0}^{‴}:={S_0}^{\prime }$ and ${S_i}^{‴}:={S_i}^{\prime }\setminus int{S_{i-1}}^{\prime }$ for $0<i$, which is closed on ${S_i}^{\prime }$ (and on $T_{\alpha }$), and as a closed subset of the compact subspace, by the proposition that any closed subset of any compact topological space is compact, is compact on ${S_i}^{\prime }$, and is compact on $T_{\alpha }$ by the proposition that for any topological space, any compact subset of any subspace is compact on the base space. ${S_i}^{‴}$ does not intersect any ${S_j}^{‴}$ where $j<i-1$ or $i+1<j$, because while ${S_i}^{‴}={S_i}^{\prime }\setminus int{S_{i-1}}^{\prime }$ and ${S_j}^{‴}={S_j}^{\prime }\setminus int{S_{j-1}}^{\prime }$, when $j<i-1$, ${S_j}^{\prime }\subseteq int{S_{i-1}}^{\prime }$, and when $i+1<j$, the roles of $i,j$ can be replaced. $\{{S_i}^{‴}|i\in J\}$ covers $T_{\alpha }$, because for any $p\in T_{\alpha }$, $p\in {S_i}^{\prime }$ for an $i$, and if $p\notin int{S_{i-1}}^{\prime }$, $p\in {S_i}^{‴}$, otherwise, $p\in {S_{i-1}}^{\prime }$, if $p\notin int{S_{i-2}}^{\prime }$, $p\in {S_{i-1}}^{‴}$, otherwise, $p\in {S_{i-2}}^{\prime }$, and so on, after all, at least $p\in {S_0}^{\prime }={S_0}^{‴}$.


By the proposition that any subspace of any Hausdorff topological space is Hausdorff and the proposition that any compact Hausdorff topological space is normal, ${S_i}^{‴}$ is normal. For ${S_i}^{‴}$, ${S_i}^{\prime }\setminus int{S_i}^{\prime }$ is closed on ${S_i}^{‴}$, because $int{S_i}^{\prime }$ is open on $T_{\alpha }$, $int{S_i}^{\prime }\cap {S_i}^{‴}$ is open on ${S_i}^{‴}$, ${S_i}^{\prime }\setminus int{S_i}^{\prime }={S_i}^{‴}\setminus (int{S_i}^{\prime }\cap {S_i}^{‴})$. For any $0<i$, ${S_{i-1}}^{\prime }\setminus int{S_{i-1}}^{\prime }$ is closed on ${S_i}^{‴}$, because ${S_{i-1}}^{\prime }\setminus int{S_{i-1}}^{\prime }={S_{i-1}}^{\prime }\cap {S_i}^{‴}$, but ${S_{i-1}}^{\prime }$ is closed on $T_{\alpha }$, and ${S_{i-1}}^{\prime }\cap {S_i}^{‴}$ is closed on ${S_i}^{‴}$. $({S_i}^{\prime }\setminus int{S_i}^{\prime })\cap ({S_{i-1}}^{\prime }\setminus int{S_{i-1}}^{\prime })=\mathrm{\varnothing }$, because ${S_{i-1}}^{\prime }\subseteq int{S_i}^{\prime }$.

Let us take take $U_{0,1}=\mathrm{\varnothing }$. If ${S_0}^{\prime }\setminus int{S_0}^{\prime }\ne \mathrm{\varnothing }$, let us take$U_{0,2}={S_0}^{‴}$. If ${S_0}^{\prime }\setminus int{S_0}^{\prime }=\mathrm{\varnothing }$, let us take $U_{0,1}=\mathrm{\varnothing }$.

These 5 paragraphs are for $0<i$.

If ${S_i}^{\prime }\setminus int{S_i}^{\prime }\ne \mathrm{\varnothing }$ and ${S_{i-1}}^{\prime }\setminus int{S_{i-1}}^{\prime }\ne \mathrm{\varnothing }$, as ${S_i}^{‴}$ is normal, there are some disjoint open (on ${S_i}^{‴}$) subsets, $U_{i,1}\subseteq {S_i}^{‴}$ and $U_{i,2}\subseteq {S_i}^{‴}$ such that ${S_{i-1}}^{\prime }\setminus int{S_{i-1}}^{\prime }\subseteq U_{i,1}$ and ${S_i}^{\prime }\setminus int{S_i}^{\prime }\subseteq U_{i,2}$ and $U_{i,1}\cap U_{i,2}=\mathrm{\varnothing }$.

If ${S_i}^{\prime }\setminus int{S_i}^{\prime }\ne \mathrm{\varnothing }$ and ${S_{i-1}}^{\prime }\setminus int{S_{i-1}}^{\prime }=\mathrm{\varnothing }$, take $U_{i,1}=\mathrm{\varnothing }$, open on ${S_i}^{‴}$, and $U_{i,2}={S_i}^{‴}$, open on ${S_i}^{‴}$.

If ${S_i}^{\prime }\setminus int{S_i}^{\prime }=\mathrm{\varnothing }$ and ${S_{i-1}}^{\prime }\setminus int{S_{i-1}}^{\prime }\ne \mathrm{\varnothing }$, take $U_{i,1}={S_i}^{‴}$, open on ${S_i}^{‴}$ and $U_{i,2}=\mathrm{\varnothing }$, open on ${S_i}^{‴}$.

If ${S_i}^{\prime }\setminus int{S_i}^{\prime }=\mathrm{\varnothing }$ and ${S_{i-1}}^{\prime }\setminus int{S_{i-1}}^{\prime }=\mathrm{\varnothing }$, take $U_{i,1}=\mathrm{\varnothing }$, open on ${S_i}^{‴}$ and $U_{i,2}=\mathrm{\varnothing }$, open on ${S_i}^{‴}$.

In each case, ${S_{i-1}}^{\prime }\setminus int{S_{i-1}}^{\prime }\subseteq U_{i,1}$ and ${S_i}^{\prime }\setminus int{S_i}^{\prime }\subseteq U_{i,2}$ and $U_{i,1}\cap U_{i,2}=\mathrm{\varnothing }$.

Let us define ${S_0}^{⁗}:={S_0}^{‴}\cup U_{1,1}$ and ${S_i}^{⁗}:={S_i}^{‴}\cup U_{i-1,2}\cup U_{i+1,1}$ for $0<i$. Anyway, ${S_i}^{⁗}$ is open on $T_{\alpha }$, which will be proved in the next 6 paragraphs.


${S_0}^{⁗}=U_{1,1}\cup int{S_0}^{\prime }$ and ${S_i}^{⁗}=(U_{i+1,1}\cup (int{S_i}^{\prime }\setminus S_{i-1}^{\prime }))\cup (U_{i-1,2}\cup (int{S_i}^{\prime }\setminus S_{i-1}^{\prime }))$.

if $U_{1,1}$ is not empty, around any point, $p\in U_{1,1}$, there is an open (on ${S_1}^{‴}$) neighborhood, $N_p\subseteq U_{1,1}$; $N_p={N_p}^{\prime }\cap {S_1}^{‴}$ where ${N_p}^{\prime }$ is an open neighborhood of $p$ on $T_{\alpha }$, ${N_p}^{″}:={N_p}^{\prime }\cap int{S_1}^{\prime }$ can be used instead because ${N_p}^{″}\cap {S_1}^{‴}=N_p$ anyway because $N_p\subseteq U_{1,1}\subseteq int{S_1}^{\prime }$ (because $U_{1,1}\subseteq {S_1}^{‴}\setminus U_{1,2}$ and ${S_1}^{\prime }\setminus int{S_1}^{\prime }\subseteq U_{1,2}$, so, $U_{1,1}\subseteq {S_1}^{‴}\setminus ({S_1}^{\prime }\setminus int{S_1}^{\prime })=({S_1}^{\prime }\setminus int{S_0}^{\prime })\setminus ({S_1}^{\prime }\setminus int{S_1}^{\prime })=int{S_1}^{\prime }\setminus int{S_0}^{\prime }\subseteq int{S_1}^{\prime }$), but ${N_p}^{″}\subseteq U_{1,1}\cup int{S_0}^{\prime }$, because otherwise, as ${N_p}^{″}\subseteq int{S_1}^{\prime }$, if there was a point, $p^{\prime }\in {N_p}^{″}$ such that $p^{\prime }\in int{S_1}^{\prime }\setminus (U_{1,1}\cup int{S_0}^{\prime })$, which would mean that $p^{\prime }\notin U_{1,1}$, $p^{\prime }\in {S_1}^{‴}={S_1}^{\prime }\setminus int{S_0}^{\prime }$, so, $p^{\prime }\in {N_p}^{″}\cap {S_1}^{‴}=N_p\subseteq U_{1,1}$, a contradiction against $p^{\prime }\notin U_{1,1}$; around any point, $p\in int{S_0}^{\prime }$, there is an open (on $T_{\alpha }$) neighborhood, $N_p\subseteq int{S_0}^{\prime }$, because $int{S_0}^{\prime }$ is open on $T_{\alpha }$; so, $U_{1,1}\cup int{S_0}^{\prime }$ is open on $T_{\alpha }$.

These 2 paragraphs are for $0<i$.

$U_{i+1,1}\cup (int{S_i}^{\prime }\setminus S_{i-1}^{\prime })=(U_{i+1,1}\cup int{S_i}^{\prime })\setminus S_{i-1}^{\prime }$, because ${S_{i-1}}^{\prime }$ does not intersect $U_{i+1,1}$ (because ${S_{i-1}}^{\prime }\subseteq int{S_i}^{\prime }$ and $U_{i+1,1}\subseteq {S_{i+1}}^{‴}={S_{i+1}}^{\prime }\setminus int{S_i}^{\prime }$), and if $U_{i+1,1}$ is not empty, around any point, $p\in U_{i+1,1}$, there is an open (on ${S_{i+1}}^{‴}$) neighborhood, $N_p\subseteq U_{i+1,1}$; $N_p={N_p}^{\prime }\cap {S_{i+1}}^{‴}$ where ${N_p}^{\prime }$ is an open neighborhood of $p$ on $T_{\alpha }$, ${N_p}^{″}:={N_p}^{\prime }\cap int{S_{i+1}}^{\prime }$ can be used instead because ${N_p}^{″}\cap {S_{i+1}}^{‴}=N_p$ anyway because $N_p\subseteq U_{i+1,1}\subseteq int{S_{i+1}}^{\prime }$ (because $U_{i+1,1}\subseteq {S_{i+1}}^{‴}\setminus U_{i+1,2}$ and ${S_{i+1}}^{\prime }\setminus int{S_{i+1}}^{\prime }\subseteq U_{i+1,2}$, so, $U_{i+1,1}\subseteq {S_{i+1}}^{‴}\setminus ({S_{i+1}}^{\prime }\setminus int{S_{i+1}}^{\prime })=({S_{i+1}}^{\prime }\setminus int{S_i}^{\prime })\setminus ({S_{i+1}}^{\prime }\setminus int{S_{i+1}}^{\prime })=int{S_{i+1}}^{\prime }\setminus int{S_i}^{\prime }\subseteq int{S_{i+1}}^{\prime }$), but ${N_p}^{″}\subseteq U_{i+1,1}\cup int{S_i}^{\prime }$, because otherwise, as ${N_p}^{″}\subseteq int{S_{i+1}}^{\prime }$, if there was a point, $p^{\prime }\in {N_p}^{″}$ such that $p^{\prime }\in int{S_{i+1}}^{\prime }\setminus (U_{i+1,1}\cup int{S_i}^{\prime })$, which would mean that $p^{\prime }\notin U_{i+1,1}$, $p^{\prime }\in {S_{i+1}}^{‴}={S_{i+1}}^{\prime }\setminus int{S_i}^{\prime }$, so, $p^{\prime }\in {N_p}^{″}\cap {S_{i+1}}^{‴}=N_p\subseteq U_{i+1,1}$, a contradiction against $p^{\prime }\notin U_{i+1,1}$; around any point, $p\in int{S_i}^{\prime }$, there is an open (on $T_{\alpha }$) neighborhood, $N_p\subseteq int{S_i}^{\prime }$, because $int{S_i}^{\prime }$ is open on $T_{\alpha }$; so, $U_{i+1,1}\cup int{S_i}^{\prime }$ is open on $T_{\alpha }$, and $(U_{i+1,1}\cup int{S_i}^{\prime })\setminus S_{i-1}^{\prime }$ is open on $T_{\alpha }$ by the proposition that any open set minus any closed set is open.

As for $U_{i-1,2}\cup (int{S_i}^{\prime }\setminus S_{i-1}^{\prime })$, if $U_{i-1,2}$ is not empty, around any point, $p\in U_{i-1,2}$, there is an open (on ${S_{i-1}}^{‴}$) neighborhood, $N_p\subseteq U_{i-1,2}$, $N_p={N_p}^{\prime }\cap {S_{i-1}}^{‴}$ where ${N_p}^{\prime }$ is open on $T_{\alpha }$, ${N_p}^{″}={N_p}^{\prime }\cap (int{S_i}^{\prime }\setminus {S_{i-2}}^{\prime })$ can be used instead because ${N_p}^{″}\cap {S_{i-1}}^{‴}=N_p$ anyway because $N_p\subseteq U_{i-1,2}\subseteq int{S_i}^{\prime }\setminus {S_{i-2}}^{\prime }$ (because $U_{i-1,2}\subseteq S_{i-1}^{‴}\setminus U_{i-1,1}$ and ${S_{i-2}}^{\prime }\setminus int{S_{i-2}}^{\prime }\subseteq U_{i-1,1}$, so, $U_{i-1,2}\subseteq {S_{i-1}}^{‴}\setminus ({S_{i-2}}^{\prime }\setminus int{S_{i-2}}^{\prime })=({S_{i-1}}^{\prime }\setminus int{S_{i-2}}^{\prime })\setminus ({S_{i-2}}^{\prime }\setminus int{S_{i-2}}^{\prime })={S_{i-1}}^{\prime }\setminus {S_{i-2}}^{\prime }\subseteq int{S_i}^{\prime }\setminus {S_{i-2}}^{\prime }$, as ${S_{i-1}}^{\prime }\setminus int{S_i}^{\prime }$), but ${N_p}^{″}\subseteq U_{i-1,2}\cup (int{S_i}^{\prime }\setminus S_{i-1}^{\prime })$, because otherwise, as ${N_p}^{″}\subseteq int{S_i}^{\prime }\setminus {S_{i-2}}^{\prime }$, if there was a point, $p^{\prime }\in {N_p}^{″}$ such that $p^{\prime }\in (int{S_i}^{\prime }\setminus {S_{i-2}}^{\prime })\setminus (U_{i-1,2}\cup (int{S_i}^{\prime }\setminus S_{i-1}^{\prime }))$, which would mean that $p^{\prime }\notin U_{i-1,2}$, $p^{\prime }\in {S_{i-1}}^{‴}={S_{i-1}}^{\prime }\setminus int{S_{i-2}}^{\prime }$, because $p^{\prime }\in {S_{i-1}}^{\prime }$ (because otherwise, $p^{\prime }\in int{S_i}^{\prime }\setminus {S_{i-1}}^{\prime }$, then, $p^{\prime }\notin (int{S_i}^{\prime }\setminus {S_{i-2}}^{\prime })\setminus (U_{i-1,2}\cup (int{S_i}^{\prime }\setminus S_{i-1}^{\prime }))$, a contradiction) and $p^{\prime }\notin {S_{i-2}}^{\prime }$, so, $p^{\prime }\in {N_p}^{″}\cap {S_{i-1}}^{‴}=N_p\subseteq U_{i-1,2}$, a contradiction against $p^{\prime }\notin U_{i-1,2}$; for any point, $p\in int{S_i}^{\prime }\setminus S_{i-1}^{\prime }$, there is an open (on $T_{\alpha }$) neighborhood, $N_p\subseteq int{S_i}^{\prime }\setminus S_{i-1}^{\prime }$, because $int{S_i}^{\prime }\setminus S_{i-1}^{\prime }$ is open on $T_{\alpha }$, by the proposition that any open set minus any closed set is open, so, $U_{i-1,2}\cup (int{S_i}^{\prime }\setminus S_{i-1}^{\prime })$ is open on $T_{\alpha }$.

So, ${S_i}^{⁗}$ is open on $T_{\alpha }$.

${S_i}^{⁗}$ does not intersect any ${S_j}^{⁗}$ where $j<i-1$ or $i+1<j$, because for $j=i-2$, if $0<i-2$, ${S_{i-2}}^{⁗}={S_{i-2}}^{‴}\cup U_{i-3,2}\cup U_{i-1,1}$ and ${S_i}^{⁗}={S_i}^{‴}\cup U_{i-1,2}\cup U_{i+1,1}$, ${S_{i-2}}^{‴}\cap {S_i}^{‴}=\mathrm{\varnothing }$ and $U_{i-1,1}\cap U_{i-1,2}=\mathrm{\varnothing }$ and the other combinations are obviously empty; if $0=i-2$, just $U_{i-3,2}$ does not exist and the claim holds; for $j<i-2$, it is more obvious; for $i+1<j$, the roles of $i,j$ can be exchanged.

Now, for any open cover, $\{V_{\beta }|\beta \in B\}$ where $B$ is a possibly uncountable indices set, of $T_{\alpha }$, it is an open cover of each ${S_i}^{‴}$, and as ${S_i}^{‴}$ is compact, there is a finite subcover, $\{V_{i,j}|j\in J_i\}$ where $J_i$ is a finite indices set that depends on $i$, and let us take $\{{V_{i,j}}^{\prime }:=V_{i,j}\cap {S_i}^{⁗}|j\in J_i\}$, which is an open cover of ${S_i}^{‴}$, contained in ${S_i}^{⁗}$.

${\cup }_i\{{V_{i,j}}^{\prime }|j\in J_i\}$ is a locally finite refinement of the original open cover, because it is an open cover because $T_{\alpha }$ is covered by $\{{S_i}^{‴}\}$ and each ${S_i}^{‴}$ is covered by it; it is a refinement because ${V_{i,j}}^{\prime }\subseteq V_{i,j}=V_{\beta }$; it is locally finite because for any point, $p\in T_{\alpha }$, $p\in {S_i}^{‴}$ for an $i$, ${S_i}^{⁗}$ is an open neighborhood of $p$, which intersects only ${S_{i-1}}^{⁗}$ and ${S_{i+1}}^{⁗}$, and so, intersects only finite ${V_{i,j}}^{\prime }$s contained in ${S_{i-1}}^{⁗}$, ${S_i}^{⁗}$, and ${S_{i+1}}^{⁗}$.

So, each $T_{\alpha }$ is paracompact, and $T=\{T_{\alpha }|\alpha \in A\}$ is paracompact.

Let us suppose that $T$ is paracompact.

Let us take an open cover of $T$ by taking around each point, $p\in T$, an open neighborhood whose closure is compact, which is possible by the proposition that for any locally compact Hausdorff topological space, around any point, there is an open neighborhood whose closure is compact. As $T$ is paracompact, there is a locally finite refinement, while the closure of any new open neighborhood is contained in the compact closure of the corresponding original neighborhood, and as a closed set in the compact subspace, the new closure is compact on the original closure and is compact on $T$ by the proposition that for any topological space, any compact subset of any subspace is compact on the base space. Let us denote this refinement $\{U_{\beta }|\beta \in B\}$ where $B$ is a possibly uncountable indices set.

Let us inductively define a sequence of open sets, $\{V_{\beta ,i}\}$, whose closures are compact, beginning from each $U_{\beta }$. $V_{\beta ,0}:=U_{\beta }$; for $V_{\beta ,i}$ for $0<i$, take all the $U_{\beta }$s that intersect $\overline{V_{\beta ,i-1}}$, whose number is finite by the proposition that for any locally finite cover of any topological space, any compact subset intersects only finite elements of the cover. Those $U_{\beta }$s form an open cover of $\overline{V_{\beta ,i-1}}$, and let us define $V_{\beta ,i}$ as the union of those $U_{\beta }$s. $\overline{V_{\beta ,i}}$ is the union of the closures of those $U_{\beta }$s by the proposition that the closure of the union of any finite number of subsets is the union of the closures of the subsets, and is compact by the proposition that for any topological space, the union of any finite compact subsets is compact. $\overline{V_{\beta ,i}}\subseteq V_{\beta ,i+1}$.

$V_{\beta }:={\cup }_i{V}_{\beta ,i}$, which is the union of some countable $U_{\beta }$s. $\overline{V_{\beta }}$ is the union of the closures of those $U_{\beta }$s by the proposition that for any locally finite open cover of any topological space, the closure of the union of any possibly uncountable open sets in the cover is the union of the closures of the open sets, but each closure is contained in a $\overline{V_{\beta ,i}}$, but $\overline{V_{\beta ,i}}\subseteq V_{\beta ,i+1}\subseteq V_{\beta }$, so, $\overline{V_{\beta }}=V_{\beta }$. $V_{\beta }$ is $\sigma$-compact.

$\{V_{\beta }|\beta \in B\}$ covers $T$, because for any point, $p\in T$, there is a $U_{\beta }$ such that $p\in U_{\beta }$, and $p\in V_{\beta }$.

Let us prove that for any $V_{\alpha },V_{\beta }$ such that $\alpha ,\beta \in B$, $V_{\alpha }=V_{\beta }$ or $V_{\alpha }\cap V_{\beta }=\mathrm{\varnothing }$..

Let us suppose that $V_{\alpha }\cap V_{\beta }\ne \mathrm{\varnothing }$. There are a point, $p\in V_{\alpha }\cap V_{\beta }$, and a $U_{\gamma }$ such that $\gamma \in B,p\in U_{\gamma }$. $U_{\gamma }\subseteq V_{\alpha }$, because $U_{\gamma }\cap \overline{V_{\alpha ,i}}\ne \mathrm{\varnothing }$ for an $i$, so, $U_{\gamma }\subseteq V_{\alpha ,i+1}\subseteq V_{\alpha }$. $V_{\gamma }\subseteq V_{\alpha }$, because otherwise, $V_{\gamma }\cap (T\setminus V_{\alpha })\ne \mathrm{\varnothing }$; there would be an $i$ such that $V_{\gamma ,i}\subseteq V_{\alpha }$ (so, $\overline{V_{\gamma ,i}}\subseteq V_{\alpha }$ as $V_{\alpha }$ is closed) and a $\delta \in B$ such that $U_{\delta }\cap \overline{V_{\gamma ,i}}\ne \mathrm{\varnothing }$ and $U_{\delta }\cap (T\setminus V_{\alpha })\ne \mathrm{\varnothing }$, because $V_{\gamma ,0}\subseteq V_{\alpha }$ and $V_{\gamma ,i}$ would have to get out of $V_{\alpha }$ at an $i$, but $U_{\delta }\cap \overline{V_{\alpha ,j}}\ne \mathrm{\varnothing }$ for a $j$, because $\overline{V_{\gamma ,i}}\subseteq V_{\alpha }$ and $U_{\delta }\cap \overline{V_{\gamma ,i}}\ne \mathrm{\varnothing }$, then, $U_{\delta }\subseteq V_{\alpha ,j+1}\subseteq V_{\alpha }$, a contradiction against $U_{\delta }\cap (T\setminus V_{\alpha })\ne \mathrm{\varnothing }$.


But in fact, $V_{\gamma }=V_{\alpha }$, as is proven as follows.

Let us suppose that $V_{\gamma }\subset V_{\alpha }$.

$V_{\alpha }$ is generated from $U_{\alpha }$ and $U_{\alpha }\subseteq V_{\alpha }$, but $U_{\alpha }\subseteq V_{\gamma }$ or $U_{\alpha }\subseteq V_{\alpha }\setminus V_{\gamma }$, because otherwise, $\mathrm{\neg }{U}_{\alpha }\subseteq V_{\gamma }$ and $U_{\alpha }\cap V_{\gamma }\ne \mathrm{\varnothing }$, then, as $U_{\alpha }\cap \overline{V_{\gamma ,i}}\ne \mathrm{\varnothing }$ for an $i$, $U_{\alpha }\subseteq V_{\gamma ,i+1}\subseteq V_{\gamma }$, a contradiction.

If $U_{\alpha }\subseteq V_{\gamma }$, $V_{\alpha ,i}\subseteq V_{\gamma }$ for an $i$ (because at least $V_{\alpha ,0}\subseteq V_{\gamma }$), $\overline{V_{\alpha ,i}}\subseteq V_{\gamma }$ because $V_{\gamma }$ is closed, then, there would be a $U_{\beta }$ such that $\overline{V_{\alpha ,i}}\cap U_{\beta }\ne \mathrm{\varnothing }$ and $U_{\beta }\cap (V_{\alpha }\setminus V_{\gamma })\ne \mathrm{\varnothing }$ for an $i$ in order for $V_{\alpha ,i+1}$ not to keep being inside $V_{\gamma }$, but then, $U_{\beta }\cap V_{\gamma }\ne \mathrm{\varnothing }$ as $\overline{V_{\alpha ,i}}\subseteq V_{\gamma }$, $\overline{V_{\gamma ,j}}\cap U_{\beta }\ne \mathrm{\varnothing }$ for a $j$, so, $U_{\beta }\subseteq V_{\gamma ,j+1}\subseteq V_{\gamma }$, a contradiction against $U_{\beta }\cap (V_{\alpha }\setminus V_{\gamma })\ne \mathrm{\varnothing }$, so, $U_{\alpha }\subseteq V_{\gamma }$ is impossible.


If $U_{\alpha }\subseteq V_{\alpha }\setminus V_{\gamma }$, $V_{\alpha ,i}\cap V_{\gamma }=\mathrm{\varnothing }$ for an $i$ (because at least $V_{\alpha ,0}\cap V_{\gamma }=\mathrm{\varnothing }$), so, $\overline{V_{\alpha ,i}}\cap V_{\gamma }=\mathrm{\varnothing }$ by the proposition that for any disjoint subset and open set, the closure of the subset and the open set are disjoint, and $V_{\alpha ,i+1}\cap V_{\gamma }=\mathrm{\varnothing }$, because otherwise, there would be a $U_{\beta }$ such that $\overline{V_{\alpha ,i}}\cap U_{\beta }\ne \mathrm{\varnothing }$ and $U_{\beta }\cap V_{\gamma }\ne \mathrm{\varnothing }$, but then, $U_{\beta }\cap V_{\gamma ,j}\ne \mathrm{\varnothing }$ for a $j$, $U_{\beta }\subseteq V_{\gamma ,j+1}\subseteq V_{\gamma }$, which would contradict $\overline{V_{\alpha ,i}}\cap V_{\gamma }=\mathrm{\varnothing }$ and $\overline{V_{\alpha ,i}}\cap U_{\beta }\ne \mathrm{\varnothing }$, which would mean that there would be a $p\in \overline{V_{\alpha ,i}}\cap U_{\beta }$ such that $p\notin V_{\gamma }$, so, $p\in U_{\beta }$ and $p\notin V_{\gamma }$, which would contradict $U_{\beta }\subseteq V_{\gamma }$. So, $V_{\alpha }\cap V_{\gamma }=\mathrm{\varnothing }$, which contradicts that $V_{\gamma }\subset V_{\alpha }$, so, $U_{\alpha }\subseteq V_{\alpha }\setminus V_{\gamma }$ is impossible.


So, $V_{\gamma }\subset V_{\alpha }$ was wrong, and $V_{\gamma }=V_{\alpha }$

Likewise, $V_{\gamma }=V_{\beta }$.

So, $V_{\alpha }=V_{\beta }$.

So, $\{V_{\beta }|\beta \in B\}$ with the duplications removed, $\{T_{\alpha }|\alpha \in A\subseteq B\}$ where $T_{\alpha }=V_{\beta }$ for a $\beta \in B$, is a disjoint open cover of $T$, and $T={\cup }_{\alpha }{T}_{\alpha }$ is the union of some disjoint open $\sigma$-compact subspaces.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>