description/proof of that for topological space, point, and neighborhood of point, neighborhood of point on neighborhood is neighborhood of point on base space
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of neighborhood of 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 topological space, any point, and any neighborhood of the point, any neighborhood of the point on the neighborhood is a neighborhood of the point on the base space.
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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(p\): \(\in T\)
\(N'_p\): \(\in \{\text{ the neighborhoods of } p \text{ on } T\}\)
\(N_p\): \(\in \{\text{ the neighborhoods of } p \text{ on } N'_p\}\)
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Statements:
\(N_p\): \(\in \{\text{ the neighborhoods of } p \text{ on } T\}\)
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2: Note
This proposition is less obvious than that any open neighborhood of the point on any open neighborhood is an open neighborhood of the point on the base space.
3: Proof
Whole Strategy: Step 1: take an open neighborhood of \(p\) on \(T\), \(U'_p \subseteq T\), such that \(U'_p \subseteq N'_p\); Step 2: take an open neighborhood of \(p\) on \(N'_p\), \(U_p \subseteq N'_p\), such that \(U_p \subseteq N_p\) and an open neighborhood of \(p\) on \(T\), \(U''_p \subseteq T\), such that \(U_p = U''_p \cap N'_p\); Step 3: see that \(U'_p \cap U''_p \subseteq T\) is an open neighborhood of \(p\) on \(T\) such that \(U'_p \cap U''_p \subseteq N_p\).
Step 1:
As \(N'_p\) is a neighborhood of \(p\) on \(T\), there is an open neighborhood of \(p\) on \(T\), \(U'_p \subseteq T\), such that \(p \in U'_p \subseteq N'_p\).
Step 2:
As \(N_p\) is a neighborhood of \(p\) on \(N'_p\), there is an open neighborhood of \(p\) on \(N'_p\), \(U_p \subseteq N'_p\), such that \(U_p \subseteq N_p\).
According to the definition of subspace topology, there is an open neighborhood of \(p\) on \(T\), \(U''_p \subseteq T\), such that \(U_p = U''_p \cap N'_p\).
Step 3:
Let us see that \(U'_p \cap U''_p \subseteq T\) is an open neighborhood of \(p\) on \(T\) such that \(U'_p \cap U''_p \subseteq N_p\).
\(p \in U'_p \cap U''_p\).
\(U'_p \cap U''_p\) is an open subset of \(T\) as a finite intersection of open subsets.
So, \(U'_p \cap U''_p\) is an open neighborhood of \(p\) on \(T\).
\(U'_p \cap U''_p \subseteq N'_p \cap U''_p = U_p \subseteq N_p\).
So, \(N_p\) satisfies the requirement to be an open neighborhood of \(p\) on \(T\).
