A description/proof of that induced functional structure on topological subspace by inclusion is functional structure
Topics
About: topological space
The table of contents of this article
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.
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.
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.
Main Body
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, \(\{F_{T_2} (U = U' \cap T_2) = \{f: U \rightarrow \mathbb{R}\vert f \text{ is continuous } \land \text{ for any point, } p \in U, \text{ there are an open neighborhood, } U'_p \subseteq U', \text{and a } g \in F_{T_1} (U'_p) \text{ such that } f\vert_{U'_p \cap T_2} = g\vert_{U'_p \cap T_2}\}\}\), is a functional structure.
2: Proof
1st, let us prove that it is well-defined, which means that while \(U'\) is not unique, it does not depend on the choice of \(U'\). 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 \(U'\) satisfy the requirement for \(U''\)? \(U''_p = U'_p \cap U''\) can be taken instead of \(U'_p\). Then, \(U''_p \subseteq U''\); \(g \vert_{U''_p} \in F_{T_1} (U''_p)\) and \(f\vert_{U''_p \cap T_2} = g\vert_{U''_p \cap T_2}\). 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 \(U'_p \subseteq U'\) and \(g \in F_{T_2} (U'_p)\) such that \((f_1 + f_2) \vert_{U'_p \cap T_2} = g\vert_{U'_p \cap T_2}\)? There are \(U'_{p, i} \subseteq U'\) and \(g_i \in F_{T_1} (U'_{p, i})\) such that \(f_i\vert_{U'_{p, i} \cap T_2} = g_i\vert_{U'_{p, i} \cap T_2}\). Let us define \(U'_p := U'_{p, 1} \cap U'_{p, 2}\) and \(g := g_1\vert_{U'_p} + g_2\vert_{U'_p}\). Then, \(U'_p\) and \(g\) will satisfy the conditions, because \(g_i \vert_{U'_p} \in F_{T_1} (U'_p)\) and \(g \in F_{T_1} (U'_p)\); \((f_1 + f_2) \vert_{U'_p \cap T_2} = (g_1 + g_2) \vert_{U'_p \cap T_2} = g \vert_{U'_p \cap T_2}\). Let us check that the subset is closed under scalar multiplications. Let us suppose that \(f \in F_{T_2} (U)\) and \(r \in \mathbb{R}\). \(r f \in F_{T_2} (U)\)? \(r f\) is continuous. Are there \(U'_p \subseteq U'\) and \(g \in F_{T_2} (U'_p)\) such that \(r f\vert_{U'_p \cap T_2} = g\vert_{U'_p \cap T_2}\)? There are \(U'_p \subseteq U'\) and \(g \in F_{T_1} (U'_p)\) such that \(f\vert_{U'_p \cap T_2} = g\vert_{U'_p \cap T_2}\). \(r g \in F_{T_1} (U'_p)\) and \(r f\vert_{U'_p \cap T_2} = r g\vert_{U'_p \cap T_2}\). 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 \(U'_p \subseteq U'\) and \(g \in F_{T_2} (U'_p)\) such that \((f_1 f_2) \vert_{U'_p \cap T_2} = g\vert_{U'_p \cap T_2}\)? There are \(U'_{p, i} \subseteq U'\) and \(g_i \in F_{T_1} (U'_{p, i})\) such that \(f_i\vert_{U'_{p, i} \cap T_2} = g_i\vert_{U'_{p, i} \cap T_2}\). Let us define \(U'_p := U'_{p, 1} \cap U'_{p, 2}\) and \(g := g_1\vert_{U'_p} g_2\vert_{U'_p}\). Then, \(U'_p\) and \(g\) will satisfy the conditions, because \(g_i \vert_{U'_p} \in F_{T_1} (U'_p)\) and \(g \in F_{T_1} (U'_p)\); \((f_1 f_2) \vert_{U'_p \cap T_2} = (g_1 g_2) \vert_{U'_p \cap T_2} = g \vert_{U'_p \cap T_2}\). 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, \(f: U \rightarrow \mathbb{R}\), there are \(U'_p = U'\) and the constant function, \(g \in F_{T_1} (U'_p)\), such that \(f \vert_{U'_p \cap T_2} = g \vert_{U'_p \cap T_2}\).
Let us prove that the "induced functional structure" satisfies the 3rd condition: for any open \(V \subseteq U\) and any \(f \in F_{T_2} (U)\), \(f\vert_V \in F_{T_2} (V)\). \(V = V' \cap T_2\) where \(V' \subseteq U'\) is open. For any point, \(p \in V \subseteq U\), there are \(U'_p \subseteq U'\) and \(g \in F_{T_1} (U'_p)\) such that \(f\vert_{U'_p \cap T_2} = g\vert_{U'_p \cap T_2}\). Let us define \(U''_p := U'_p \cap V'\). Then, \(g\vert_{U''_p} \in F_{T_1} (U''_p)\), and \(f\vert_{U''_p \cap T_2} = g\vert_{U''_p \cap T_2}\).
Let us prove that the "induced functional structure" satisfies the 4th condition: for any open cover \(U = \cup_\alpha U_\alpha\) and any continuous \(f: U \rightarrow \mathbb{R}\) such that \(f_{U_\alpha} \in F_{T_2} (U_\alpha)\), \(f \in F_{T_2} (U)\). \(U_\alpha = U'_\alpha \cap T_2\) where \(U'_\alpha\) can be taken to be \(U'_\alpha \subseteq U'\), because if not, \(U'_\alpha \cap U'\) can be used instead. For any point, \(p \in U\), \(p \in U_\alpha = U'_\alpha \cap T_2\) for an \(\alpha\). There are a \(U'_{p, \alpha} \subseteq U'_\alpha \subseteq U'\) and a \(g \in F_{T_1} (U'_{p, \alpha})\) such that \(f\vert_{U'_{p, \alpha} \cap T_2} = g\vert_{U'_{p, \alpha} \cap T_2}\). So, \(U'_p := U'_{p, \alpha}\) and \(g\) suffices.
3: Note
Just calling \(\{F_{T_2} (U = U' \cap T_2) = \{f: U \rightarrow \mathbb{R}\vert f \text{ is continuous } \land \text{ for any point, } p \in U, \text{ there are an open neighborhood, } U'_p \subseteq U', \text{and a } g \in F_{T_1} (U'_p) \text{ such that } f\vert_{U'_p \cap T_2} = g\vert_{U'_p \cap T_2}\}\}\) "induced functional structure" does not guarantee that \(\{F_{T_2} (U)\}\) is a functional structure, which requires satisfying the 4 conditions.
