528: For Infinite Product Topological Space and Closed Subset, Point on Product Space Whose Each Finite-Components-Projection Belongs to Corresponding Projection of Subset Belongs to Subset

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description/proof of that for infinite product topological space and closed subset, point on product space whose each finite-components-projection belongs to corresponding projection of subset belongs to subset

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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 1
• 4: Proof
• 5: Note 2

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Starting Context

• The reader knows a definition of product topological space.

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Target Context

• The reader will have a description and a proof of the proposition that for any infinite product topological space and any closed subset, any point on the product space whose each finite-components-projection belongs to the corresponding projection of the subset belongs to the subset.

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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:
$A$: $\in \{\text{ the possibly uncountable infinite index sets }\}$
$\{T_{\alpha }|\alpha \in A\}$: $T_{\alpha }\in \{\text{ the topological spaces }\}$
$T$: $={\times }_{\alpha \in A}{T}_{\alpha }$ with the product topology
$C$: $\subseteq T$, $\in \{\text{ the closed subsets }\}$
$p$: $\in T$
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Statements:
$\mathrm{\forall }J\subset A,J\in \{\text{ the finite index sets }\}({\pi }_J(p)\in {\pi }_J(C))$, where ${\pi }_J:T\to {\times }_{j\in J}{T}_j$ is the projection
$⟹$
$p\in C$
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2: Natural Language Description

For any possibly uncountable infinite index set, $A$, any topological spaces, $\{T_{\alpha }|\alpha \in A\}$, the product topological space, $T:={\times }_{\alpha \in A}{T}_{\alpha }$, any closed $C\subseteq T$, and any $p\in T$, ${\pi }_J(p)\in {\pi }_J(C)$ for each $J\subset A$, where $J$ is a finite index set and ${\pi }_J:T\to {\times }_{j\in J}{T}_j$ is the projection, implies that $p\in C$.

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3: Note 1

A typical case is $T=T_1\times T_2\times ...$, a infinitely countable product, and ${\pi }_J(p)\in {\pi }_J(C)$ where $J=\{1,...,k\}$ for each $k$ implies $p\in C$: the requirement for such $J$s implies the requirement for each $J^{\prime }\subseteq J$, so, implies the requirement for each finite $J^{\prime }\subseteq A$.

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4: Proof

Let us suppose that ${\pi }_J(p)\in {\pi }_J(C)$ for each $J$.

$T\setminus C\subseteq T$ is open.

Let us suppose that $p\notin C$.

$p\in T\setminus C$. There would be an open neighborhood, $U_p\subseteq T\setminus C$, of $p$. $U_p={\cup }_{\beta \in B}{\times }_{\alpha \in A}{U}_{\beta ,\alpha }$, where $B$ would be a possibly uncountable index set and $U_{\beta ,\alpha }\subseteq T_{\alpha }$ would be open while only finite number of $U_{\beta ,\alpha }$ s would not be $T_{\alpha }$ s for each $\beta$ (see Note of the definition of product topology).

There would be a $\beta$ such that $p\in {\times }_{\alpha \in A}{U}_{\beta ,\alpha }\subseteq T\setminus C$. As only finite of $U_{\beta ,\alpha }$ s would not be $T_{\alpha }$ s, let $J$ be of the finite components. ${\pi }_J(p)\in {\pi }_J({\times }_{\alpha \in A}{U}_{\beta ,\alpha })={\times }_{j\in J}{U}_{\beta ,j}$. Then, there would be a $p^{\prime }\in C$ such that ${\pi }_J(p^{\prime })={\pi }_J(p)\in {\times }_{j\in J}{U}_{\beta ,j}$. But $p^{\prime }\in {\times }_{\alpha \in A}{U}_{\beta ,\alpha }$, because $U_{\beta ,\alpha }=T_{\alpha }$ for each $\alpha \notin J$ and so, $p^{\prime \alpha }\in U_{\beta ,\alpha }$ when $\alpha \in J$ and when $\alpha \in A\setminus J$.

So, $p^{\prime }\in {\times }_{\alpha \in A}{U}_{\beta ,\alpha }\subseteq {\cup }_{\beta \in B}{\times }_{\alpha \in A}{U}_{\beta ,\alpha }=U_p\subseteq T\setminus C$, which would mean $p^{\prime }\notin C$, a contradiction.

So, $p\in C$.

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5: Note 2

When $C\subseteq T$ is not closed, $p\in C$ is not necessarily implied as is proved in another article.

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References

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