803: For Topological Space and Locally Finite Set of Closed Subsets, Union of Set Is Closed

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description/proof of that for topological space and locally finite set of closed subsets, union of set is closed

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Natural Language Description
• 3: Note
• 4: Proof

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Starting Context

• The reader knows a definition of locally finite set of subsets of topological space.
• The reader knows a definition of closed set.
• The reader knows a definition of union of set.
• The reader admits the local criterion for openness.

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Target Context

• The reader will have a description and a proof of the proposition that for any topological space and any locally finite set of closed subsets, the union of the set is closed.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$T$: $\in \{\text{ the topological spaces }\}$
$A$: $\in \{\text{ the possibly uncountable index sets }\}$
$\{C_{\alpha }|\alpha \in A\}$: $\in \{\text{ the locally finite sets of closed subsets of }T\}$
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Statements:
$\cup \{C_{\alpha }|\alpha \in A\}\subseteq T\in \{\text{ the closed subsets }\}$
//

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2: Natural Language Description

For any topological space, $T$, any possibly uncountable index set, $A$, and any locally finite set of closed subsets of $T$, $\{C_{\alpha }|\alpha \in A\}$, the union of $\{C_{\alpha }|\alpha \in A\}$, $\cup \{C_{\alpha }|\alpha \in A\}\subseteq T$, is closed.

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3: Note

When $A$ is finite, of course, $\cup \{C_{\alpha }|\alpha \in A\}\subseteq T$ is closed; this proposition is about a sufficient condition with which the union of some infinite number of closed subsets is closed.

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4: Proof

Whole Strategy: prove that $T\setminus \cup \{C_{\alpha }|\alpha \in A\}$ is open by the local criterion for openness; Step 1: let $t\in T\setminus \cup \{C_{\alpha }|\alpha \in A\}$ be any and take any open neighborhood of $t$, $U_t\subseteq T$, that intersects only finite of $C_{\alpha }$s, $\{C_{\alpha }|\alpha \in B\}$; Step 2: take $U_t^{\prime }:=U_t\cap {\cap }_{\alpha \in B}(T\setminus C_{\alpha })\subseteq T$ and see that $U_t^{\prime }$ is an open neighborhood of $t$ on $T$; Step 3: see that $U_t^{\prime }\subseteq T\setminus \cup \{C_{\alpha }|\alpha \in A\}$; Step 4: conclude the proposition.

Step 1:

Let $t\in T\setminus \cup \{C_{\alpha }|\alpha \in A\}$ be any.

Let us take any open neighborhood of $t$, $U_t\subseteq T$, such that there is a finite subset (may be empty), $B\subseteq A$, such that for each $\alpha \in B$, $U_t\cap C_{\alpha }\ne \mathrm{\varnothing }$ and for each $\alpha \in A\setminus B$, $U_t\cap C_{\alpha }=\mathrm{\varnothing }$, which is possible by the definition of locally finite set of subsets of topological space.

Step 2:

When $B$ is empty, let us take $U_t^{\prime }:=U_t$. $U_t^{\prime }$ is an open neighborhood of $t$ on $T$.

Let us suppose that $B$ is not empty for the rest of this step.

Let us take $U_t^{\prime }:=U_t\cap {\cap }_{\alpha \in B}(T\setminus C_{\alpha })\subseteq T$.

$t\in U_t^{\prime }$, because $t\in U_t$ and for each $\alpha \in B$, $t\in T\setminus C_{\alpha }$, because $t\in T\setminus \cup \{C_{\alpha }|\alpha \in A\}$, which means that $t\notin C_{\alpha }$ for each $\alpha \in A$, which means that $t\notin C_{\alpha }$ for each $\alpha \in B\subseteq A$.

As $T\setminus C_{\alpha }\subseteq T$ is open and $B$ is finite, $U_t^{\prime }$ is open on $T$ as the finite intersection of open subsets.

So, $U_t^{\prime }$ is an open neighborhood of $t$ on $T$.

Step 3:

Let us see that $U_t^{\prime }\subseteq T\setminus \cup \{C_{\alpha }|\alpha \in A\}$.

Let $t^{\prime }\in U_t^{\prime }$ be any. $t^{\prime }\in T\setminus C_{\alpha }$ for each $\alpha \in B$, which means that $t^{\prime }\notin C_{\alpha }$. $t^{\prime }\in U_t$, which means that $t^{\prime }\notin C_{\alpha }$ for each $\alpha \in A\setminus B$ because $U_t\cap C_{\alpha }=\mathrm{\varnothing }$. So, $t^{\prime }\notin C_{\alpha }$ for each $\alpha \in A$. So, $t^{\prime }\notin \cup \{C_{\alpha }|\alpha \in A\}$, which implies that $t^{\prime }\in T\setminus \cup \{C_{\alpha }|\alpha \in A\}$.

So, $U_t^{\prime }\subseteq T\setminus \cup \{C_{\alpha }|\alpha \in A\}$.

Step 4:

By the local criterion for openness, $T\setminus \cup \{C_{\alpha }|\alpha \in A\}$ is open on $T$, so, $\cup \{C_{\alpha }|\alpha \in A\}$ is closed on $T$.

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References

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