374: Induced Functional Structure on Topological Subspace by Inclusion Is Functional Structure

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A description/proof of that induced functional structure on topological subspace by inclusion is functional structure

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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 functional structure on topological space.
• The reader knows a definition of induced functional structure on topological subspace by inclusion.

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Target Context

• The reader will have a description and a proof of the proposition that the induced functional structure on any topological subspace of any topological space by the inclusion is a functional structure.

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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 topological space, $T_1$, any topological subspace, $T_2\subseteq T_1$, and any functional structure, $\{F_{T_1}(U)\}$, the induced functional structure, $\{F_{T_2}(U=U^{\prime }\cap T_2)=\{f:U\to \mathbb{R}|f\text{ is continuous }\wedge \text{ for any point, }p\in U,\text{ there are an open neighborhood, }{U}_p^{\prime }\subseteq U^{\prime },\text{and a }g\in F_{T_1}(U_p^{\prime })\text{ such that }f{|}_{U_p^{\prime }\cap T_2}=g{|}_{U_p^{\prime }\cap T_2}\}\}$, is a functional structure.

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2: Proof

1st, let us prove that it is well-defined, which means that while $U^{\prime }$ is not unique, it does not depend on the choice of $U^{\prime }$. Let us suppose that there is another open subset, $U^{″}\subseteq T_1$, such that $U=U^{″}\cap T_2$. Does any $f$ that satisfies the requirement for $U^{\prime }$ satisfy the requirement for $U^{″}$? $U_p^{″}=U_p^{\prime }\cap U^{″}$ can be taken instead of $U_p^{\prime }$. Then, $U_p^{″}\subseteq U^{″}$; $g{|}_{U_p^{″}}\in F_{T_1}(U_p^{″})$ and $f{|}_{U_p^{″}\cap T_2}=g{|}_{U_p^{″}\cap T_2}$. So, yes, $f$ satisfies the requirement for $U^{″}$.

Let us prove that the "induced functional structure" satisfies the 1st condition: $F_{T_2}(U)$ is a sub-algebra of the all-the-continuous functions algebra. $F_{T_2}(U)$ is a subset of the all-the-continuous functions algebra, because $f$ is continuous by the definition. Let us check that the subset is closed under addition. Let us suppose that $f_i\in F_{T_2}(U)$. $f_1+f_2\in F_{T_2}(U)$? $f_1+f_2$ is continuous. Are there $U_p^{\prime }\subseteq U^{\prime }$ and $g\in F_{T_2}(U_p^{\prime })$ such that $(f_1+f_2){|}_{U_p^{\prime }\cap T_2}=g{|}_{U_p^{\prime }\cap T_2}$? There are $U_{p,i}^{\prime }\subseteq U^{\prime }$ and $g_i\in F_{T_1}(U_{p,i}^{\prime })$ such that $f_i{|}_{U_{p,i}^{\prime }\cap T_2}=g_i{|}_{U_{p,i}^{\prime }\cap T_2}$. Let us define $U_p^{\prime }:=U_{p,1}^{\prime }\cap U_{p,2}^{\prime }$ and $g:=g_1{|}_{U_p^{\prime }}+g_2{|}_{U_p^{\prime }}$. Then, $U_p^{\prime }$ and $g$ will satisfy the conditions, because $g_i{|}_{U_p^{\prime }}\in F_{T_1}(U_p^{\prime })$ and $g\in F_{T_1}(U_p^{\prime })$; $(f_1+f_2){|}_{U_p^{\prime }\cap T_2}=(g_1+g_2){|}_{U_p^{\prime }\cap T_2}=g{|}_{U_p^{\prime }\cap T_2}$. Let us check that the subset is closed under scalar multiplications. Let us suppose that $f\in F_{T_2}(U)$ and $r\in \mathbb{R}$. $rf\in F_{T_2}(U)$? $rf$ is continuous. Are there $U_p^{\prime }\subseteq U^{\prime }$ and $g\in F_{T_2}(U_p^{\prime })$ such that $rf{|}_{U_p^{\prime }\cap T_2}=g{|}_{U_p^{\prime }\cap T_2}$? There are $U_p^{\prime }\subseteq U^{\prime }$ and $g\in F_{T_1}(U_p^{\prime })$ such that $f{|}_{U_p^{\prime }\cap T_2}=g{|}_{U_p^{\prime }\cap T_2}$. $rg\in F_{T_1}(U_p^{\prime })$ and $rf{|}_{U_p^{\prime }\cap T_2}=rg{|}_{U_p^{\prime }\cap T_2}$. Let us check that the subset is closed under multiplications. Let us suppose that $f_i\in F_{T_2}(U)$. $f_1{f}_2\in F_{T_2}(U)$? $f_1{f}_2$ is continuous. Are there $U_p^{\prime }\subseteq U^{\prime }$ and $g\in F_{T_2}(U_p^{\prime })$ such that $(f_1{f}_2){|}_{U_p^{\prime }\cap T_2}=g{|}_{U_p^{\prime }\cap T_2}$? There are $U_{p,i}^{\prime }\subseteq U^{\prime }$ and $g_i\in F_{T_1}(U_{p,i}^{\prime })$ such that $f_i{|}_{U_{p,i}^{\prime }\cap T_2}=g_i{|}_{U_{p,i}^{\prime }\cap T_2}$. Let us define $U_p^{\prime }:=U_{p,1}^{\prime }\cap U_{p,2}^{\prime }$ and $g:=g_1{|}_{U_p^{\prime }}{g}_2{|}_{U_p^{\prime }}$. Then, $U_p^{\prime }$ and $g$ will satisfy the conditions, because $g_i{|}_{U_p^{\prime }}\in F_{T_1}(U_p^{\prime })$ and $g\in F_{T_1}(U_p^{\prime })$; $(f_1{f}_2){|}_{U_p^{\prime }\cap T_2}=(g_1{g}_2){|}_{U_p^{\prime }\cap T_2}=g{|}_{U_p^{\prime }\cap T_2}$. For any $f_i\in F_{T_2}(U)$, the left distributability, $(f_1+f_2)f_3=f_1{f}_3+f_2{f}_3$, holds. For any $f_i\in F_{T_2}(U)$, the right distributability, $f_3(f_1+f_2)=f_3{f}_1+f_3{f}_1$, holds. For any $f_i\in F_{T_2}(U)$ and $r_i\in \mathbb{R}$, the compatibility with scalars, $(r_1{f}_1)(r_2{f}_2)=(r_1{r}_2)(f_1{f}_2)$, holds.

Let us prove that the "induced functional structure" satisfies the 2nd condition: $F_{T_2}(U)$ contains all the constant functions. For any constant function, $f:U\to \mathbb{R}$, there are $U_p^{\prime }=U^{\prime }$ and the constant function, $g\in F_{T_1}(U_p^{\prime })$, such that $f{|}_{U_p^{\prime }\cap T_2}=g{|}_{U_p^{\prime }\cap T_2}$.

Let us prove that the "induced functional structure" satisfies the 3rd condition: for any open $V\subseteq U$ and any $f\in F_{T_2}(U)$, $f{|}_V\in F_{T_2}(V)$. $V=V^{\prime }\cap T_2$ where $V^{\prime }\subseteq U^{\prime }$ is open. For any point, $p\in V\subseteq U$, there are $U_p^{\prime }\subseteq U^{\prime }$ and $g\in F_{T_1}(U_p^{\prime })$ such that $f{|}_{U_p^{\prime }\cap T_2}=g{|}_{U_p^{\prime }\cap T_2}$. Let us define $U_p^{″}:=U_p^{\prime }\cap V^{\prime }$. Then, $g{|}_{U_p^{″}}\in F_{T_1}(U_p^{″})$, and $f{|}_{U_p^{″}\cap T_2}=g{|}_{U_p^{″}\cap T_2}$.

Let us prove that the "induced functional structure" satisfies the 4th condition: for any open cover $U={\cup }_{\alpha }{U}_{\alpha }$ and any continuous $f:U\to \mathbb{R}$ such that $f_{U_{\alpha }}\in F_{T_2}(U_{\alpha })$, $f\in F_{T_2}(U)$. $U_{\alpha }=U_{\alpha }^{\prime }\cap T_2$ where $U_{\alpha }^{\prime }$ can be taken to be $U_{\alpha }^{\prime }\subseteq U^{\prime }$, because if not, $U_{\alpha }^{\prime }\cap U^{\prime }$ can be used instead. For any point, $p\in U$, $p\in U_{\alpha }=U_{\alpha }^{\prime }\cap T_2$ for an $\alpha$. There are a $U_{p,\alpha }^{\prime }\subseteq U_{\alpha }^{\prime }\subseteq U^{\prime }$ and a $g\in F_{T_1}(U_{p,\alpha }^{\prime })$ such that $f{|}_{U_{p,\alpha }^{\prime }\cap T_2}=g{|}_{U_{p,\alpha }^{\prime }\cap T_2}$. So, $U_p^{\prime }:=U_{p,\alpha }^{\prime }$ and $g$ suffices.

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3: Note

Just calling $\{F_{T_2}(U=U^{\prime }\cap T_2)=\{f:U\to \mathbb{R}|f\text{ is continuous }\wedge \text{ for any point, }p\in U,\text{ there are an open neighborhood, }{U}_p^{\prime }\subseteq U^{\prime },\text{and a }g\in F_{T_1}(U_p^{\prime })\text{ such that }f{|}_{U_p^{\prime }\cap T_2}=g{|}_{U_p^{\prime }\cap T_2}\}\}$ "induced functional structure" does not guarantee that $\{F_{T_2}(U)\}$ is a functional structure, which requires satisfying the 4 conditions.

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References

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