423: For Finite-Product Topological Space, Product of Neighborhoods Is Neighborhood

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A description/proof of that for finite-product topological space, product of neighborhoods is neighborhood

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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: Description
• 2: Proof
• 3: Note

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Starting Context

• The reader knows a definition of product topology.
• The reader knows a definition of neighborhood of point.

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Target Context

• The reader will have a description and a proof of the proposition that for any finite-product topological space, the product of any constituent neighborhoods is a neighborhood.

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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: Description

For any finite number of topological spaces, $T_1,T_2,...,T_n$, the product, $T=T_1\times T_2\times ...\times T_n$, any point, $p=(p^1,p^2,...,p^n)\in T$, and any neighborhoods, $\{N_{p^j}\subseteq T_j|j\in \{1,2,...,n\}\}$, of $\{p^j|j\in \{1,2,...,n\}\}$, $N_p=N_{p^1}\times N_{p^2}\times ...\times N_{p^n}\subseteq T$ is a neighborhood of $p$.

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2: Proof

Certainly, $p\in N_p$.

For each $j$, there is an open neighborhood, $U_{p^j}\subseteq T_j$, of $p^j$ such that $U_{p^j}\subseteq N_{p^j}$, by the definition of neighborhood of point. $U_p:=U_{p^1}\times U_{p^2}\times ...\times U_{p^n}\subseteq N_p$. $p\in U_p$, and $U_p$ is an open neighborhood of $p$, by the definition of product topology.

So, $N_p$ is a neighborhood of $p$.

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3: Note

For an infinite-product topological space, the logic does not apply, because $U_p$ would not be any open subset of $T$, because ${\times }_{\alpha \in A}{U}_{p^{\alpha }}$ would not be open unless only each of some finite of $U_{p^{\alpha }}$ s is different from $T_{\alpha }$.

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References

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