766: For Continuous Map from Product Topological Space into Topological Space, Induced Map with Set of Components of Domain Fixed Is Continuous

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description/proof of that for continuous map from product topological space into topological space, induced map with set of components of domain fixed is continuous

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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 1
• 2: Natural Language Description 1
• 3: Proof 1
• 4: Structured Description 2
• 5: Natural Language Description 2
• 6: Proof 2

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Starting Context

• The reader knows a definition of product topology.
• The reader knows a definition of continuous map.

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Target Context

• The reader will have a description and a proof of the proposition that for any continuous map from any product topological space into any topological space, the induced map with any set of some components of the domain fixed is continuous.

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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 1

Here is the rules of Structured Description.

Entities:
$A^{\prime }$: $\in \{\text{ the possibly uncountable index sets }\}$
$\{T_{\alpha }|\alpha \in A^{\prime }\}$: $T_{\alpha }\in \{\text{ the topological spaces }\}$
$T_1^{\prime }$: $={\times }_{\alpha \in A^{\prime }}{T}_{\alpha }$, $=\text{ the product topological space }$
$T_2^{\prime }$: $\in \{\text{ the topological spaces }\}$
$f^{\prime }$: $:T_1^{\prime }\to T_2^{\prime }$, $\in \{\text{ the continuous maps }\}$
$A^{″}$: $\subset A^{\prime }$
$T_1^{″}$: $={\times }_{\alpha \in A^{″}}{T}_{\alpha }$, $=\text{ the product topological space }$
$t_{\alpha }$: $\in T_{\alpha }$ for each $\alpha \in A^{\prime }\setminus A^{″}$
${\pi }_{A^{″}}$: $:T_1^{\prime }\to T_1^{″}$, which takes the $A^{″}$ components
${\tau }_{A^{″}}$: $:T_1^{″}\to T_1^{\prime }$, which adds the $A^{\prime }\setminus A^{″}$ components as $t_{\alpha }$ s
$f^{″}$: $:T_1^{″}\to T_2^{\prime },t^{″}\mapsto f^{\prime }({\tau }_{A^{″}}(t^{″}))$
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Statements:
$f^{″}\in \{\text{ the continuous maps }\}$
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2: Natural Language Description 1

For any possibly uncountable index set, $A^{\prime }$, any set of topological spaces, $\{T_{\alpha }|\alpha \in A^{\prime }\}$, the product topological space, $T_1^{\prime }={\times }_{\alpha \in A^{\prime }}{T}_{\alpha }$, any topological space, $T_2^{\prime }$, any continuous map, $f^{\prime }:T_1^{\prime }\to T_2^{\prime }$, any subset, $A^{″}\subset A^{\prime }$, the product topological space, $T_1^{″}={\times }_{\alpha \in A^{″}}{T}_{\alpha }$, any $t_{\alpha }\in T_{\alpha }$ for each $\alpha \in A^{\prime }\setminus A^{″}$, the map, ${\pi }_{A^{″}}:T_1^{\prime }\to T_1^{″}$, which takes the $A^{″}$ components, the map, ${\tau }_{A^{″}}:T_1^{″}\to T_1^{\prime }$, which adds the $A^{\prime }\setminus A^{″}$ components as $t_{\alpha }$ s, and $f^{″}:T_1^{″}\to T_2^{\prime },t^{″}\mapsto f^{\prime }({\tau }_{A^{\prime }}(t^{″}))$, $f^{‴}$ is a continuous map.

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3: Proof 1

Whole Strategy: Step 1: for each point, $p^{″}\in T_1^{″}$, take $p^{\prime }:={\tau }_{A^{″}}(p^{″})\in T_1^{\prime }$, and for each neighborhood of $f^{\prime }(p^{\prime })$, $N_{f^{\prime }(p^{\prime })}^{\prime }$, take an open neighborhood of $p^{\prime }$, ${\times }_{\alpha \in A^{\prime }}{U}_{\alpha ,\beta }\subseteq T_1^{\prime }$, such that $f^{\prime }({\times }_{\alpha \in A^{\prime }}{U}_{\alpha ,\beta })\subseteq N_{f^{\prime }(p^{\prime })}^{\prime }$; Step 2: see that ${\times }_{\alpha \in A^{″}}{U}_{\alpha ,\beta }\subseteq T_1^{″}$ is an open neighborhood of $p^{″}$ such that $f^{″}({\times }_{\alpha \in A^{″}}{U}_{\alpha ,\beta })\subseteq N_{f^{\prime }(p^{\prime })}^{\prime }$.

Step 1:

Let $p^{″}\in T_1^{″}$ be any.

Let $p^{\prime }:={\tau }_{A^{″}}(p^{″})\in T_1^{\prime }$.

For any neighborhood of $f^{\prime }(p^{\prime })\in T_2^{\prime }$, $N_{f^{\prime }(p^{\prime })}^{\prime }\subseteq T_2^{\prime }$, there is an open neighborhood of $p^{\prime }$, $U_{p^{\prime }}^{\prime }\subseteq T_1^{\prime }$, such that $f^{\prime }(U_{p^{\prime }}^{\prime })\subseteq N_{f^{\prime }(p^{\prime })}^{\prime }$, because $f^{\prime }$ is continuous.

$U_{p^{\prime }}^{\prime }={\cup }_{\beta \in B}{\times }_{\alpha \in A^{\prime }}{U}_{\alpha ,\beta }$ where $B$ is any possibly uncountable index set and $U_{\alpha ,\beta }$ is an open subset of $T_{\alpha }$ with each of only finite of $U_{\alpha ,\beta }$ s for each $\beta$ is not $T_{\alpha }$, by Note for the definition of product topology. $p^{\prime }\in {\times }_{\alpha \in A^{\prime }}{U}_{\alpha ,\beta }$ for a $\beta$, and we can take only that single ${\times }_{\alpha \in A^{\prime }}{U}_{\alpha ,\beta }$, because that is an open neighborhood of $p^{\prime }$ and satisfies $f({\times }_{\alpha \in A^{\prime }}{U}_{\alpha ,\beta })\subseteq N_{f^{\prime }(p^{\prime })}^{\prime }$.

${\times }_{\alpha \in A^{″}}{U}_{\alpha ,\beta }$ is an open neighborhood of $p^{″}$ and $f^{″}({\times }_{\alpha \in A^{″}}{U}_{\alpha ,\beta })\subseteq f^{\prime }({\times }_{\alpha \in A^{\prime }}{U}_{\alpha ,\beta })\subseteq N_{f^{\prime }(p^{\prime })}^{\prime }$, because for each $q\in f^{″}({\times }_{\alpha \in A^{″}}{U}_{\alpha ,\beta })$, $q=f^{″}(q^{″})$ for a $q^{″}\in {\times }_{\alpha \in A^{″}}{U}_{\alpha ,\beta }$, but ${\tau }_{A^{″}}(p^{″})\in {\times }_{\alpha \in A^{\prime }}{U}_{\alpha ,\beta }$, and $p^{″}=f^{″}(q^{″})=f^{\prime }({\tau }_{A^{\prime }}(q^{″}))\in f^{\prime }({\times }_{\alpha \in A^{\prime }}{U}_{\alpha ,\beta })$.

So, $f^{″}$ is continuous at any point, $p^{″}\in T_1^{″}$, and so, is continuous all over.

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4: Structured Description 2

Here is the rules of Structured Description.

Entities:
$J^{\prime }$: $=\{1,...,n\}$
$\{T_j|j\in J^{\prime }\}$: $T_j\in \{\text{ the topological spaces }\}$
$T_1^{\prime }$: $=T_1\times ...\times T_n$, $=\text{ the product topological space }$
$T_2^{\prime }$: $\in \{\text{ the topological spaces }\}$
$f^{\prime }$: $:T_1^{\prime }\to T_2^{\prime }$, $\in \{\text{ the continuous maps }\}$
$J^{″}$: $\subset J^{\prime }$
$T_1^{″}$: $={\times }_{j\in J^{″}}{T}_j$, $=\text{ the product topological space }$
$t_j$: $\in T_j$ for each $j\in J^{\prime }\setminus J^{″}$
${\pi }_{J^{″}}$: $:T_1^{\prime }\to T_1^{″}$, which takes the $J^{″}$ components
${\tau }_{J^{″}}$: $:T_1^{″}\to T_1^{\prime }$, which adds the $J^{\prime }\setminus J^{″}$ components as $t_j$ s
$f^{″}$: $:T_1^{″}\to T_2^{\prime },t^{″}\mapsto f^{\prime }({\tau }_{J^{″}}(t^{″}))$
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Statements:
$f^{″}\in \{\text{ the continuous maps }\}$
//

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5: Natural Language Description 2

For $J^{\prime }=\{1,...,n\}$, any set of topological spaces, $\{T_j|j\in J^{\prime }\}$, the product topological space, $T_1^{\prime }=T_1\times ...\times T_n$, any topological space, $T_2^{\prime }$, any continuous map, $f^{\prime }:T_1^{\prime }\to T_2^{\prime }$, any subset, $J^{″}\subset J^{\prime }$, the product topological space, $T_1^{″}={\times }_{j\in J^{″}}{T}_j$, any $t_j\in T_j$ for each $j\in J^{\prime }\setminus J^{″}$, the map, ${\pi }_{J^{″}}:T_1^{\prime }\to T_1^{″}$, which takes the $J^{″}$ components, ${\tau }_{J^{″}}:T_1^{″}\to T_1^{\prime }$, which adds the $J^{\prime }\setminus J^{″}$ components as $t_j$ s, and $f^{″}:T_1^{″}\to T_2^{\prime },t^{″}\mapsto f^{\prime }({\tau }_{J^{″}}(t^{″}))$, $f^{″}$ is a continuous map.

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6: Proof 2

Whole Strategy: Step 1: for each point, $p^{″}\in T_1^{″}$, take $p^{\prime }:={\tau }_{J^{″}}(p^{″})\in T_1^{\prime }$, and for each neighborhood of $f^{\prime }(p^{\prime })$, $N_{f^{\prime }(p^{\prime })}^{\prime }$, take an open neighborhood of $p^{\prime }$, $U_{1,\beta }\times ...\times U_{n,\beta }\subseteq T_1^{\prime }$, such that $f^{\prime }(U_{1,\beta }\times ...\times U_{n,\beta })\subseteq N_{f^{\prime }(p^{\prime })}^{\prime }$; Step 2: see that ${\times }_{j\in J^{″}}{U}_{j,\beta }\subseteq T_1^{″}$ is an open neighborhood of $p^{″}$ such that $f^{″}({\times }_{j\in J^{″}}{U}_{j,\beta })\subseteq N_{f^{\prime }(p^{\prime })}^{\prime }$.

Step 1:

Let $p^{″}\in T_1^{″}$ be any.

Let $p^{\prime }:={\tau }_{J^{″}}(p^{″})\in T_1^{\prime }$.

For any neighborhood of $f^{\prime }(p^{\prime })\in T_2^{\prime }$, $N_{f^{\prime }(p^{\prime })}^{\prime }\subseteq T_2^{\prime }$, there is an open neighborhood of $p^{\prime }$, $U_{p^{\prime }}^{\prime }\subseteq T_1^{\prime }$, such that $f^{\prime }(U_{p^{\prime }}^{\prime })\subseteq N_{f^{\prime }(p^{\prime })}^{\prime }$, because $f^{\prime }$ is continuous.

$U_{p^{\prime }}^{\prime }={\cup }_{\beta \in B}(U_{1,\beta }\times ...\times U_{n,\beta })$ where $B$ is any possibly uncountable index set and $U_{j,\beta }$ is an open subset of $T_j$, by Note for the definition of product topology. $p^{\prime }\in U_{1,\beta }\times ...\times U_{n,\beta }$ for a $\beta$, and we can take only that single $U_{1,\beta }\times ...\times U_{n,\beta }$, because that is an open neighborhood of $p^{\prime }$ and satisfies $f^{\prime }(U_{1,\beta }\times ...\times U_{n,\beta })\subseteq N_{f^{\prime }(p^{\prime })}^{\prime }$.

Step 2:

${\times }_{j\in J^{″}}{U}_{j,\beta }$ is an open neighborhood of $p^{″}$ and $f^{″}({\times }_{j\in J^{″}}{U}_{j,\beta })\subseteq f^{\prime }(U_{1,\beta }\times ...\times U_{n,\beta })\subseteq N_{f^{\prime }(p^{\prime })}^{\prime }$, because for each $q\in f^{″}({\times }_{j\in J^{″}}{U}_{j,\beta })$, there is a $q^{″}\in {\times }_{j\in J^{″}}{U}_{j,\beta }$ such that $q=f^{″}(q^{″})$, and ${\tau }_{J^{\prime }}(q^{″})\in U_{1,\beta }\times ...\times U_{n,\beta }$, and $q=f^{″}(q^{″})=f^{\prime }({\tau }_{J^{″}}(q^{″}))\in f^{\prime }(U_{1,\beta }\times ...\times U_{n,\beta })$.

So, $f^{″}$ is continuous at any point, $p^{″}\in T_j^{″}$, and so, is continuous all over.

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References

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