425: For Product Topological Space, Projection of Compact Subset Is Compact

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A description/proof of that for product topological space, projection of compact subset is 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 1
• 2: Proof 1
• 3: Description 2
• 4: Proof 2

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Starting Context

• The reader knows a definition of product topology.
• The reader knows a definition of compact subset.

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Target Context

• The reader will have a description and a proof of the proposition that for any product topological space, the projection of any compact subset into any constituent topological space is a compact subset of the constituent topological space.

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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 1

For any possibly uncountable number of indexed topological spaces, $\{T_{\alpha }|\alpha \in A\}$ where $A$ is any possibly uncountable indices set, the product, $T:={\times }_{\alpha \in A}{T}_{\alpha }$, and any compact subset, $S\subseteq T$, the projection, ${\pi }_{{\alpha }^{\prime }}(S)\subseteq T_{{\alpha }^{\prime }}$ where ${\pi }_{{\alpha }^{\prime }}:T\to T_{{\alpha }^{\prime }}$, is a compact subset of $T_{{\alpha }^{\prime }}$.

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2: Proof 1

For any $p\in T$, let the $\alpha$ component of $p$ be denoted as $p^{\alpha }$, which is ${\pi }_{\alpha }(p)=p^{\alpha }$.

Let $\{U_{\beta }\subseteq T_{{\alpha }^{\prime }}|\beta \in B\}$ be any open cover of ${\pi }_{{\alpha }^{\prime }}(S)$ where $B$ is any possibly uncountable indices set.

$\{{\times }_{\alpha \in A}{U}_{\alpha ,\beta }\subseteq T|\beta \in B\}$ where $U_{{\alpha }^{\prime },\beta }=U_{\beta }$ and $U_{\alpha ,\beta }=T_{\alpha }$ when $\alpha \ne {\alpha }^{\prime }$ is an open cover of $S$, because for any point, $p\in S$, $p^{{\alpha }^{\prime }}\in U_{\beta }=U_{{\alpha }^{\prime },\beta }$ for a $\beta \in B$, and $p^{\alpha }\in T_{\alpha }=U_{\alpha ,\beta }$ for any $\alpha \ne {\alpha }^{\prime }$ and the $\beta$, so, $p\in {\times }_{\alpha \in A}{U}_{\alpha ,\beta }$ for the $\beta$.

As $S$ is a compact subset of $T$, there is a finite open cover, $\{{\times }_{\alpha \in A}{U}_{\alpha ,\beta }|\beta \in J\}$ where $J\subseteq B$ is a finite indices set.

$\{U_{\beta }\subseteq T_{{\alpha }^{\prime }}|\beta \in J\}$ is a finite subcover, because for any point, $p^{{\alpha }^{\prime }}\in {\pi }_{{\alpha }^{\prime }}(S)$, it is the ${\alpha }^{\prime }$ component of a point, $p\in S$, and $p\in {\times }_{\alpha \in A}{U}_{\alpha ,\beta }$ for a $\beta \in J$, which means that $p^{{\alpha }^{\prime }}\in U_{{\alpha }^{\prime },\beta }=U_{\beta }$.

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3: Description 2

For any finite number of topological spaces, $T_1,T_2,...,T_n$, the product, $T=T_1\times T_2\times ...\times T_n$, and any compact subset, $S\subseteq T$, the projection, ${\pi }_{j^{\prime }}(S)\subseteq T_{j^{\prime }}$ where ${\pi }_{j^{\prime }}:T\to T_{j^{\prime }}$, is a compact subset of $T_{j^{\prime }}$.

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4: Proof 2

For any $p\in T$, let the $j$ component of $p$ be denoted as $p^j$, which is ${\pi }_j(p)=p^j$.

Let $\{U_{\beta }\subseteq T_{j^{\prime }}|\beta \in B\}$ be any open cover of ${\pi }_{j^{\prime }}(S)$ where $B$ is any possibly uncountable indices set.

$\{T_1\times ...\times T_{j^{\prime }-1}\times U_{\beta }\times T_{j^{\prime }+1}\times ...\times T_n\subseteq T|\beta \in B\}$ is an open cover of $S$, because for any point, $p\in S$, $p^{j^{\prime }}\in U_{\beta }$ for a $\beta \in B$, and $p^j\in T_j$ for any $j\ne j^{\prime }$ and the $\beta$, so, $p\in T_1\times ...\times T_{j^{\prime }-1}\times U_{\beta }\times T_{j^{\prime }+1}\times ...\times T_n$ for the $\beta$.

As $S$ is a compact subset of $T$, there is a finite open cover, $\{T_1\times ...\times T_{j^{\prime }-1}\times U_{\beta }\times T_{j^{\prime }+1}\times ...\times T_n|\beta \in J\}$ where $J\subseteq B$ is a finite indices set.

$\{U_{\beta }\subseteq T_{j^{\prime }}|\beta \in J\}$ is a finite subcover, because for any point, $p^{j^{\prime }}\in {\pi }_{j^{\prime }}(S)$, it is the $j^{\prime }$ component of a point, $p\in S$, and $p\in T_1\times ...\times T_{j^{\prime }-1}\times U_{\beta }\times T_{j^{\prime }+1}\times ...\times T_n$ for a $\beta \in J$, which means that $p^{j^{\prime }}\in U_{\beta }$.

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References

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