424: Finite Product of Locally Compact Topological Spaces Is Locally Compact

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A description/proof of that finite product of locally compact topological spaces is locally compact

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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 locally compact topological space.
• The reader knows a definition of product topology.
• The reader admits the proposition that for any finite-product topological space, the product of any constituent neighborhoods is a neighborhood.
• The reader admits the proposition that the compactness of any topological subset as a subset equals the compactness as a subspace.
• The reader admits the proposition that the product of any finite number of compact topological spaces is compact.
• The reader admits the proposition that for any possibly uncountable number of indexed topological spaces or any finite number of topological spaces and their subspaces, the product of the subspaces is the subspace of the product of the base spaces.

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Target Context

• The reader will have a description and a proof of the proposition that the product of any finite number of locally compact topological spaces is locally compact.

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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 locally compact topological spaces, $T_1,T_2,...,T_n$, the product, $T=T_1\times T_2\times ...\times T_n$, is a locally compact topological space.

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2: Proof

Let $p\in T$ be any point. Let $p^j$ be the $j$ component of $p$. Let $N_p\subseteq T$ be any neighborhood of $p$.

There is an open neighborhood, $U_p\subseteq T$, such that $U_p\subseteq N_p$. $U_p={\cup }_{\alpha \in A}{U}_{1,\alpha }\times U_{2,\alpha }\times ...\times U_{n,\alpha }$ where $A$ is a possibly uncountable indices set and $U_{j,\alpha }\subseteq T_j$ is an open subset, by the definition of product topology. We can take an $\alpha \in A$ such that $p\in U_{1,\alpha }\times U_{2,\alpha }\times ...\times U_{n,\alpha }$.

For each $j$, there is a compact neighborhood, $N_{p^j}\in T_j$, of $p^j$ such that $N_{p^j}\subseteq U_{j,\alpha }$, because $T_j$ is locally compact and $U_{j,\alpha }$ is a neighborhood of $p^j$.

$N_p^{\prime }:=N_{p^1}\times N_{p^2}\times ...\times N_{p^n}\subseteq U_{1,\alpha }\times U_{2,\alpha }\times ...\times U_{n,\alpha }\subseteq U_p\subseteq N_p\subseteq T$, which is a neighborhood of $p$, by the proposition that for any finite-product topological space, the product of any constituent neighborhoods is a neighborhood.

Each $N_{p^j}$ is a compact subspace, by the proposition that the compactness of any topological subset as a subset equals the compactness as a subspace. $N_p^{\prime }$ is a compact space, by the proposition that the product of any finite number of compact topological spaces is compact. $N_p^{\prime }$ is the subspace of $T$, by the proposition that for any possibly uncountable number of indexed topological spaces or any finite number of topological spaces and their subspaces, the product of the subspaces is the subspace of the product of the base spaces. $N_p^{\prime }$ is compact as a subset of $T$, by the proposition that the compactness of any topological subset as a subset equals the compactness as a subspace.

So, $N_p^{\prime }$ is a compact neighborhood of $p$ contained in $N_p$.

So, $T$ is locally compact.

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3: Note

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

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References

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