description/proof of that for map between topological spaces and domain point, if there are superspaces of domain and codomain, open neighborhoods of point and of point image on superspaces, and continuous map from domain neighborhood into codomain neighborhood that is restricted to original map on intersection of domain neighborhood and original domain, original map is continuous at point
Topics
About: topological space
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
- 4: Note
Starting Context
- The reader knows a definition of topological space.
- The reader knows a definition of continuous map at point.
- The reader admits the proposition that for any topological space and any sub topological space that is open on the base space, any subset of the subspace is open on the subspace if and only if it is open on the base space.
Target Context
- The reader will have a description and a proof of the proposition that for any map between any topological spaces and any domain point, if there are some superspaces of the domain and the codomain, some open neighborhoods of the point and of the point image on the superspaces, and a continuous map from the domain neighborhood into the codomain neighborhood that is restricted to the original map on the intersection of the domain neighborhood and the original domain, the original map is continuous at the point.
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: Structured Description
Here is the rules of Structured Description.
Entities:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(f\): \(: T_1 \to T_2\)
\(p\): \(\in T_1\)
//
Statements:
(
\(\exists T'_1, T'_2 \in \{\text{ the topological spaces }\} (T_1 \subseteq T'_1 \land T_1 \in \{\text{ the topological subspaces of } T'_1\} \land T_2 \subseteq T'_2 \land T_2 \in \{\text{ the topological subspaces of } T'_2\})\)
\(\land\)
\(\exists U'_p \in \{\text{ the open neighborhoods of } p \text{ on } T'_1\} \land \exists U'_{f (p)} \in \{\text{ the open neighborhoods of } f (p) \text{ on } T'_2\} \land \exists f': U'_p \to U'_{f (p)} (f' \vert_{U'_p \cap T_1} = f \vert_{U'_p \cap T_1} \land f' \in \{\text{ the continuous maps }\})\)
)
\(\implies\)
\(f \in \{\text{ the maps continuous at } p\}\)
//
2: Natural Language Description
For any topological spaces, \(T_1, T_2\), any map, \(f: T_1 \to T_2\), and any point, \(p \in T_1\), if there are some topological spaces, \(T'_1, T'_2\), such that \(T_1 \subseteq T'_1\) and \(T_1\) is a topological subspace of \(T'_1\) and \(T_2 \subseteq T'_2\) and \(T_2\) is a topological subspace of \(T'_2\), an open neighborhood, \(U'_p \subseteq T'_1\), of \(p\) on \(T'_1\) and an open neighborhood, \(U'_{f (p)} \subseteq T'_2\), of \(f (p)\) on \(T'_2\), and a continuous map, \(f': U'_p \to U'_{f (p)}\), such that \(f' \vert_{U'_p \cap T_1} = f \vert_{U'_p \cap T_1}\), \(f\) is continuous at \(p\).
3: Proof
Let us suppose that there are such \(T'_1, T'_2, U'_p, U'_{f (p)}, f'\).
For any open neighborhood, \(U_{f (p)} \subseteq T_2\), of \(f (p)\) on \(T_2\), there is an open neighborhood, \(U''_{f (p)} \subseteq T'_2\), of \(f (p)\) on \(T'_2\), such that \(U_{f (p)} = U''_{f (p)} \cap T_2\). \(U'_{f (p)} \cap U''_{f (p)}\) is open on \(T'_2\) and on \(U'_{f (p)}\), and \(f'^{-1} (U'_{f (p)} \cap U''_{f (p)})\) is open on \(U'_p\) and so, on \(T'_1\), by the proposition that for any topological space and any sub topological space that is open on the base space, any subset of the subspace is open on the subspace if and only if it is open on the base space. \(U_p := f'^{-1} (U'_{f (p)} \cap U''_{f (p)}) \cap T_1\) is open on \(T_1\).
In fact, \(U_p\) is named so because \(p \in U_p\), because \(f' (p) = f (p) \in U'_{f (p)}\) and \(f' (p) = f (p) \in U_{f (p)} = U''_{f (p)} \cap T_2\), so, \(f' (p) \in U'_{f (p)} \cap U''_{f (p)}\), so, \(p \in f'^{-1} (U'_{f (p)} \cap U''_{f (p)})\) and of course, \(p \in T_1\).
\(f (U_p) \subseteq U_{f (p)}\), because for each \(p' \in U_p\), \(f (p') = f' (p')\) because \(p' \in U'_p \cap T_1\) (as \(U_p\) is in the domain of \(f'\)), so, \(f (p') \in U'_{f (p)} \cap U''_{f (p)}\) (as \(U_p\) is in the preimage of \(U'_{f (p)} \cap U''_{f (p)}\) under \(f'\)), but as \(f\) is into \(T_2\), \(f (p') \in T_2\), so, \(f (p') \in U'_{f (p)} \cap U''_{f (p)} \cap T_2\), but as \(U_{f (p)} = U''_{f (p)} \cap T_2\), \(f (p') \in U'_{f (p)} \cap U_{f (p)} \subseteq U_{f (p)}\).
So, for each neighborhood, \(U_{f (p)}\), of \(f (p)\) on \(T_2\), there is a neighborhood, \(U_p\), of \(p\) on \(T_1\), such that \(f (U_p) \subseteq U_{f (p)}\), which is nothing but the definition of continuousness at point.
4: Note
The superspaces are usually taken to be some \(C^\infty\) manifolds, and the usual aim of bringing in the extended map is to check the continuousness of the original map with the extended map, which is expected to have a coordinates function representation, whose continuousness can be checked in norm sense, which equals the continuousness of the extended map, by the proposition that for any map between \(C^\infty\) manifolds, its continuousness in the topological sense equals its continuousness in the norm sense for the coordinates functions.
Note that the subspaces do not necessarily have any chart, because a subspace is not necessarily a \(C^\infty\) manifold and may be just a topological space, so, there may be no coordinates function representation of the map, by which the continuousness of the map could be checked.
Bringing in an arbitrary in-norm-sense continuous function between the topological subspaces that (the function) is not based on any superspaces should not guarantee the continuousness of the original map in topological sense, unless proved otherwise.
