527: For Infinite Product Topological Space and Subset, Point on Product Space Whose Each Finite-Components-Projection Belongs to Corresponding Projection of Subset Does Not Necessarily Belong to Subset

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description/proof of that for infinite product topological space and subset, point on product space whose each finite-components-projection belongs to corresponding projection of subset does not necessarily belong 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 an infinite product topological space and a subset, a point on the product space whose each finite-components-projection belongs to the corresponding projection of the subset does not necessarily belong 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
$S$: $\subseteq T$
$p$: $\in T$
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Statements:
$\mathrm{\forall }J\subset A,J\in \{\text{ the finite index sets }\}({\pi }_J(p)\in {\pi }_J(S))$ does not necessarily imply $p\in S$, where ${\pi }_J:T\to {\times }_{j\in J}{T}_j$ is the projection.
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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 $S\subseteq T$, and any $p\in T$, ${\pi }_J(p)\in {\pi }_J(S)$ 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, does not necessarily imply that $p\in S$.

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3: Note 1

A typical case is $T=T_1\times T_2\times ...$, a infinitely countable product, and ${\pi }_J(p)\in {\pi }_J(S)$ where $J=\{1,...,k\}$ for each $k$ does not imply $p\in S$.

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4: Proof

A counterexample suffices.

Let each $T_{\alpha }$ have at least 2 points, $p_{\alpha ,1},p_{\alpha ,2}\in T_{\alpha }$, let $S=T\setminus {\times }_{\alpha \in A}\{p_{\alpha ,1}\}$, and let each $\alpha$ component of $p$ be $p_{\alpha ,1}$.

$p\in T$. ${\pi }_J(p)\in {\pi }_J(S)$ for each $J$, because $S$ has at least 1 point, $p^{\prime }$, whose $J$ components are $p_{j,1}$ s ($j\in J$) but whose $\alpha$ component is $p_{\alpha ,2}$ where $\alpha \notin J$; $p^{\prime }\in S$, because $p_{\alpha ,2}\notin \{p_{\alpha ,1}\}$.

But $p\notin S$, because $p\in {\times }_{\alpha \in A}\{p_{\alpha ,1}\}$.

Letting $A$ be countable does not help. Let $A=\{1,2,...\}$, let each $T_j$ have at least 2 points, $p_{j,1},p_{j,2}\in T_j$, let $S=T\setminus (\{p_{1,1}\}\times \{p_{2,1}\}\times ...)$, and let $p$ be $(p_{1,1},p_{2,1},...)$.

$p\in T$. ${\pi }_J(p)\in {\pi }_J(S))$ for each $J$, because, for example, when $J=\{2,4\}$, $(p_{2,2},p_{2,1},p_{3,1},p_{4,1},...),(p_{1,1},p_{2,1},p_{3,2},p_{4,1},...),...\in S$, while ${\pi }_J(p)={\pi }_J((p_{2,2},p_{2,1},p_{3,1},p_{4,1},...)))=(p_{2,1},p_{4,1})$.

But $p\notin S$, because $p\in \{p_{1,1}\}\times \{p_{2,1}\}\times ...$.

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5: Note 2

When $S\subseteq T$ is closed, $p\in S$ is implied as is proved in another article.

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References

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