description/proof of that for maps between arbitrary subspaces of topological spaces continuous at corresponding points, composition is continuous at point
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of map continuous at point.
- The reader knows a definition of subspace topology of subset of topological space.
Target Context
- The reader will have a description and a proof of the proposition that for any maps between any arbitrary subspaces of any topological spaces continuous at any corresponding points, the composition 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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(T_3\): \(\in \{\text{ the topological spaces }\}\)
\(S_1\): \(\in \{\text{ the topological subspaces of } T_1\}\)
\(S_2\): \(\in \{\text{ the topological subspaces of } T_2\}\)
\(S'_2\): \(\in \{\text{ the topological subspaces of } T_2\}\), such that \(S_2 \subseteq S'_2\)
\(S_3\): \(\in \{\text{ the topological subspaces of } T_3\}\)
\(p\): \(\in S_1\)
\(f_1\): \(: S_1 \to S_2\), \(\in \{\text{ the maps continuous at } p\}\)
\(f_2\): \(: S'_2 \to S_3\), \(\in \{\text{ the maps continuous at } f_1 (p)\}\)
\(f_2 \circ f_1\): \(: S_1 \to S_3\)
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Statements:
\(f_2 \circ f_1 \in \{\text{ the maps continuous at } p\}\)
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2: Note
It is crucial that \(S_2\) and \(S'_2\) are regarded to be some subspaces of the same \(T_2\): if \(f_2\) is continuous with \(S'_2\) regarded as a topological subspace of another topological space, \(T'_2\), with the same set but with a different topology, this proposition cannot be applied.
For an obvious example, if we were allowed to choose a different topology for \(T'_2\), I would choose the discrete topology for \(T'_2\), which would make any \(f_2\) be continuous (any map from any discrete topological space is continuous), and the proposition would imply that for whatever \(f_2\), \(f_2 \circ f_1\) would be continuous, which is of course not true.
We will sometimes call a composition of continuous maps "legitimate chain of continuous maps" when the requirements for this proposition are satisfied (when \(S_2\) and \(S'_2\) are regarded to be some topological subspaces of the same \(T_2\) together with \(S_2 \subseteq S'_2\)). That is because we (or at least I) tend to slip in checking the requirements.
3: Proof
Whole Strategy: take any open neighborhood of \(f_2 \circ f_1 (p)\), \(U_{f_2 \circ f_1 (p)} \subseteq S_3 \subseteq T_3\), and an open neighborhood of \(p\), \(U_p \subseteq S_1 \subseteq T_1\), such that \(f_2 \circ f_1 (U_p) \subseteq U_{f_2 \circ f_1 (p)}\); Step 1: take any open neighborhood of \(f_2 \circ f_1 (p)\), \(U_{f_2 \circ f_1 (p)} \subseteq S_3 \subseteq T_3\); Step 2: take an open neighborhood of \(f_1 (p)\), \(U'_{f_1 (p)} \subseteq S'_2 \subseteq T_2\), such that \(f_2 (U'_{f_1 (p)}) \subseteq U_{f_2 \circ f_1 (p)}\); Step 3: take \(U_{f_1 (p)} := U'_{f_1 (p)} \cap S_2 \subseteq S_2 \subseteq T_2\) and see that it is an open neighborhood of \(f_1 (p)\) on \(S_2\); Step 4: take an open neighborhood of \(p\), \(U_p \subseteq S_1 \subseteq T_1\), such that \(f_1 (U_p) \subseteq U_{f_1 (p)}\), and see that \(f_2 \circ f_1 (U_p) \subseteq U_{f_2 \circ f_1 (p)}\).
Step 1:
Let us take any open neighborhood of \(f_2 \circ f_1 (p)\), \(U_{f_2 \circ f_1 (p)} \subseteq S_3 \subseteq T_3\).
Step 2:
As \(f_2\) is continuous at \(f_1 (p)\), there is an open neighborhood of \(f_1 (p)\), \(U'_{f_1 (p)} \subseteq S'_2 \subseteq T_2\), such that \(f_2 (U'_{f_1 (p)}) \subseteq U_{f_2 \circ f_1 (p)}\).
Step 3:
Let us take \(U_{f_1 (p)} := U'_{f_1 (p)} \cap S_2 \subseteq S_2 \subseteq T_2\).
Let us see that \(U_{f_1 (p)}\) is an open neighborhood of \(f_1 (p)\) on \(S_2\).
As \(U'_{f_1 (p)}\) is an open neighborhood of \(f_1 (p)\) on \(S'_2\), \(U'_{f_1 (p)} = U' \cap S'_2\) where \(U' \subseteq T_2\) is an open subset of \(T_2\), by the definition of subspace topology.
\(U_{f_1 (p)} = U'_{f_1 (p)} \cap S_2 = U' \cap S'_2 \cap S_2 = U' \cap S_2\), because \(S_2 \subseteq S'_2\).
That means that \(U_{f_1 (p)}\) is an open subset of \(S_2\), by the definition of subspace topology.
\(f_1 (p) \in U_{f_1 (p)}\), because \(f_1 (p) \in U'_{f_1 (p)}\) and \(f_1 (p) \in \cap S_2\).
So, \(U_{f_1 (p)}\) is an open neighborhood of \(f_1 (p)\) on \(S_2\).
Step 4:
As \(f_1\) is continuous at \(p\), there is an open neighborhood of \(p\), \(U_p \subseteq S_1 \subseteq T_1\), such that \(f_1 (U_p) \subseteq U_{f_1 (p)}\).
\(f_2 \circ f_1 (U_p) \subseteq f_2 (U_{f_1 (p)}) \subseteq f_2 (U'_{f_1 (p)}) \subseteq U_{f_2 \circ f_1 (p)}\).
