314: Product of Topological Subspaces Is Subspace of Product of Base Spaces

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A description/proof of that product of topological subspaces is subspace of product of base spaces

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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 subspace topology.

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Target Context

• The reader will have a description and a proof of 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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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, $\{T_{\alpha }^{\prime }|\alpha \in A\}$ where $A$ is any possibly uncountable indices set, and their subspaces, $\{T_{\alpha }\subseteq T_{\alpha }^{\prime }|\alpha \in A\}$, the product of the subspaces, $T={\times }_{\alpha \in A}{T}_{\alpha }$ is the subspace of the product of the base spaces, $T^{\prime }={\times }_{\alpha \in A}{T}_{\alpha }^{\prime }$.

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2: Proof 1

For any open set, $U\subseteq T$, with $T$ as the product space, $U={\cup }_{\beta \in B}{\times }_{\alpha \in A}{U}_{\alpha ,\beta }$ where $B$ is a possibly uncountable indices set and $U_{\alpha ,\beta }\subseteq T_{\alpha }$ is an open subset of $T_{\alpha }$ where only finite of $U_{\alpha ,\beta }$ s for each $\beta$ are not $T_{\alpha }$, as is described in Note in the article on a definition of product topology. $U_{\alpha ,\beta }=U_{\alpha ,\beta }^{\prime }\cap T_{\alpha }$ where $U_{\alpha ,\beta }^{\prime }\subseteq T_{\alpha }^{\prime }$ is open on $T_{\alpha }^{\prime }$ and only finite of $U_{\alpha ,\beta }^{\prime }$ s for each $\beta$ are not $T_{\alpha }^{\prime }$.

${\times }_{\alpha \in A}{U}_{\alpha ,\beta }={\times }_{\alpha \in A}(U_{\alpha ,\beta }^{\prime }\cap T_{\alpha })={\times }_{\alpha \in A}{U}_{\alpha ,\beta }^{\prime }\cap {\times }_{\alpha \in A}{T}_{\alpha }$, because for any $p\in {\times }_{\alpha \in A}(U_{\alpha ,\beta }^{\prime }\cap T_{\alpha })$, $p(\alpha )\in U_{\alpha ,\beta }^{\prime }\cap T_{\alpha }$, $p\in {\times }_{\alpha \in A}{U}_{\alpha ,\beta }^{\prime }$ and $p\in {\times }_{\alpha \in A}{T}_{\alpha }$, $p\in {\times }_{\alpha \in A}{U}_{\alpha ,\beta }^{\prime }\cap {\times }_{\alpha \in A}{T}_{\alpha }$; for any $p\in {\times }_{\alpha \in A}{U}_{\alpha ,\beta }^{\prime }\cap {\times }_{\alpha \in A}{T}_{\alpha }$, $p(\alpha )\in U_{\alpha ,\beta }^{\prime }$ and $p(\alpha )\in T_{\alpha }$, $p(\alpha )\in U_{\alpha ,\beta }^{\prime }\cap T_{\alpha }$, $p\in {\times }_{\alpha \in A}(U_{\alpha ,\beta }^{\prime }\cap T_{\alpha })$.

As ${\times }_{\alpha \in A}{U}_{\alpha ,\beta }^{\prime }$ is open on $T^{\prime }$, $U$ is an open set of $T$ as a subspace of $T^{\prime }$.

For any open set, $U\subseteq T$, with $T$ as the subspace of $T^{\prime }$, $U=U^{\prime }\cap T$ where $U^{\prime }\subseteq T^{\prime }$ is an open subset of $T^{\prime }$. $U^{\prime }={\cup }_{\beta \in B}{\times }_{\alpha \in A}{U}_{\alpha ,\beta }^{\prime }$ where $B$ is a possibly uncountable indices set and $U_{\alpha ,\beta }^{\prime }\subseteq T_{\alpha }^{\prime }$ is an open subset on $T_{\alpha }^{\prime }$ where only finite of $U_{\alpha ,\beta }^{\prime }$ s for each $\beta$ are not $T_{\alpha }^{\prime }$, as is described in Note in the article on a definition of product topology.

$U=({\cup }_{\beta \in B}{\times }_{\alpha \in A}{U}_{\alpha ,\beta }^{\prime })\cap T={\cup }_{\beta \in B}({\times }_{\alpha \in A}{U}_{\alpha ,\beta }^{\prime }\cap {\times }_{\alpha \in A}{T}_{\alpha })={\cup }_{\beta \in B}({\times }_{\alpha \in A}(U_{\alpha ,\beta }^{\prime }\cap T_{\alpha }))$, as has been shown above.

As $U_{\alpha ,\beta }^{\prime }\cap T_{\alpha }$ is open on $T_{\alpha }$ where only finite of $U_{\alpha ,\beta }^{\prime }\cap T_{\alpha }$ s are not $T_{\alpha }$ for each $\beta$, $U$ is an open subset of $T$ as the product space.

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3: Description 2

For any finite number of topological spaces, $T_1^{\prime },T_2^{\prime },...,T_n^{\prime }$, and their subspaces, $T_1\subseteq T_1^{\prime },T_2\subseteq T_2^{\prime },...,T_n\subseteq T_n^{\prime }$, the product of the subspaces, $T=T_1\times T_2\times ...\times T_n$ is the subspace of the product of the base spaces, $T^{\prime }=T_1^{\prime }\times T_2^{\prime }\times ...\times T_n^{\prime }$.

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4: Proof 2

For any open set, $U\subseteq T$, with $T$ as the product space, $U={\cup }_{\alpha \in A}{U}_{1-\alpha }\times U_{2-\alpha }\times ...\times U_{n-\alpha }$ where $A$ is a possibly uncountable indices set and $U_{i-\alpha }\subseteq T_i$ is an open set on $T_i$, as is described in Note in the article on a definition of product topology. $U_{i-\alpha }=U_{i-\alpha }^{\prime }\cap T_i$ where $U_{i-\alpha }^{\prime }\subseteq T_i^{\prime }$ is open on $T_i^{\prime }$.

$U_{1-\alpha }\times U_{2-\alpha }\times ...\times U_{n-\alpha }=(U_{1-\alpha }^{\prime }\cap T_1)\times (U_{2-\alpha }^{\prime }\cap T_2)\times ...\times (U_{n-\alpha }^{\prime }\cap T_n)=(U_{1-\alpha }^{\prime }\times U_{2-\alpha }^{\prime }\times ...\times U_{n-\alpha }^{\prime })\cap (T_1\times T_2\times ...\times T_n)$, because for any $p\in (U_{1-\alpha }^{\prime }\cap T_1)\times (U_{2-\alpha }\cap T_2)\times ...\times (U_{n-\alpha }\cap T_n)$, $p=⟨p_1,p_2,...,p_n⟩$, $p_i\in U_{i-\alpha }^{\prime }\cap T_i$, $p\in U_{1-\alpha }^{\prime }\times U_{2-\alpha }^{\prime }\times ...\times U_{n-\alpha }^{\prime }$ and $p\in T_1\times T_2\times ...\times T_n$, $p\in (U_{1-\alpha }^{\prime }\times U_{2-\alpha }^{\prime }\times ...\times U_{n-\alpha }^{\prime })\cap (T_1\times T_2\times ...\times T_n)$; for any $p\in (U_{1-\alpha }^{\prime }\times U_{2-\alpha }^{\prime }\times ...\times U_{n-\alpha }^{\prime })\cap (T_1\times T_2\times ...\times T_n)$, $p=⟨p_1,p_2,...,p_n⟩$, $p_i\in U_{i-\alpha }^{\prime }$ and $p_i\in T_i$, $p_i\in U_{i-\alpha }^{\prime }\cap T_i$, $p\in (U_{1-\alpha }^{\prime }\cap T_1)\times (U_{2-\alpha }^{\prime }\cap T_2)\times ...\times (U_{n-\alpha }^{\prime }\cap T_n)$. As $(U_{1-\alpha }^{\prime }\times U_{2-\alpha }^{\prime }\times ...\times U_{n-\alpha }^{\prime })$ is open on $T^{\prime }$, $U$ is an open set of $T$ as a subspace of $T^{\prime }$.

For any open set, $U\subseteq T$, with $T$ as the subspace of $T^{\prime }$, $U=U^{\prime }\cap T$ where $U^{\prime }\subseteq T^{\prime }$ is an open set on $T^{\prime }$. $U^{\prime }={\cup }_{\alpha \in A}{U}_{1-\alpha }^{\prime }\times U_{2-\alpha }^{\prime }\times ...\times U_{n-\alpha }^{\prime }$ where $A$ is a possibly uncountable indices set and $U_{i-\alpha }^{\prime }\subseteq T_i^{\prime }$ is an open set on $T_i^{\prime }$, as is described in Note in the article on a definition of product topology.

$U=({\cup }_{\alpha \in A}{U}_{1-\alpha }^{\prime }\times U_{2-\alpha }^{\prime }\times ...\times U_{n-\alpha }^{\prime })\cap T={\cup }_{\alpha \in A}((U_{1-\alpha }^{\prime }\times U_{2-\alpha }^{\prime }\times ...\times U_{n-\alpha }^{\prime })\cap T)$. But $(U_{1-\alpha }^{\prime }\times U_{2-\alpha }^{\prime }\times ...\times U_{n-\alpha }^{\prime })\cap T=(U_{1-\alpha }^{\prime }\cap T_1)\times (U_{2-\alpha }^{\prime }\cap T_2)\times ...\times (U_{n-\alpha }^{\prime }\cap T_n)$, as has been shown above. As $U_{i-\alpha }^{\prime }\cap T_i$ is open on $T_i$, $U$ is an open set of $T$ as the product space.

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References

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