description/proof of that for topological space, subspace, and point on subspace, intersection of neighborhoods basis at point on base space and subspace is neighborhoods basis at point on subspace
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of neighborhoods basis at point on topological space.
- The reader knows a definition of topological subspace.
- The reader admits the proposition that for any topological space and any point on any subspace, the intersection of any neighborhood of the point on the base space and the subspace is a neighborhood on the subspace.
Target Context
- The reader will have a description and a proof of the proposition that for any topological space, any subspace, and any point on the subspace, the intersection of any neighborhoods basis at the point on the base space and the subspace is a neighborhoods basis at the point on the subspace.
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:
\(T'\): \(\in \{\text{ the topological spaces }\}\)
\(T\): \(\in \{\text{ the topological subspaces of } T'\}\)
\(t\): \(\in T\)
\(B'_t\): \(\in \{\text{ the neighborhoods bases at } t \text{ on } T'\}\), \(= \{N'_{t, j} \vert j \in J\}\), where \(J\) is an index set
\(B_t\): \(= \{N'_{t, j} \cap T \vert j \in J\}\)
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Statements:
\(B_t \in \{\text{ the neighborhoods bases at } t \text{ on } T\}\)
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2: Proof
Whole Strategy: Step 1: see that each \(N'_{t, j} \cap T\) is a neighborhood of \(t\) on \(T\); Step 2: see that for each neighborhood of \(t\) on \(T\), \(N_t\), there is an \(N'_{t, j} \cap T\) such that \(N'_{t, j} \cap T \subseteq N_t\); Step 3: conclude the proposition.
Step 1:
Each \(N'_{t, j} \cap T \in B_t\) is a neighborhood of \(t\) on \(T\), by the proposition that for any topological space and any point on any subspace, the intersection of any neighborhood of the point on the base space and the subspace is a neighborhood on the subspace.
Step 2:
Let \(N_t \subseteq T\) be any neighborhood of \(t\) on \(T\).
There is an open neighborhood of \(t\) on \(T\), \(U_t \subseteq T\), such that \(U_t \subseteq N_t\).
\(U_t = U'_t \cap T\), where \(U'_t \subseteq T'\) is an open neighborhood of \(t\) on \(T'\), by the definition of subspace topology.
There is an \(N'_{t, j} \in B'_t\) such that \(N'_{t, j} \subseteq U'_t\), by the definition of neighborhoods basis at point.
\(N'_{t, j} \cap T \subseteq U'_t \cap T = U_t \subseteq N_t\).
Step 3:
So, \(B_t\) is a neighborhoods basis at \(t\) on \(T\).
