description/proof of that 2nd-countable topological space is 1st-countable
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of 1st-countable topological space.
- The reader knows a definition of 2nd-countable topological space.
Target Context
- The reader will have a description and a proof of the proposition that any 2nd-countable topological space is 1st-countable.
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:
\(T\): \(\in \{\text{ the 2nd-countable topological spaces }\}\)
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Statements:
\(T \in \{\text{ the 1st-countable topological spaces }\}\)
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2: Proof
Whole Strategy: Step 1: let \(B = \{U_j \vert j \in J\}\) be any countable basis, and for each \(t \in T\), take \(\{U_j \in B \vert t \in U_j\}\), and see that it is a countable neighborhoods basis at \(t\).
Step 1:
Let \(B = \{U_j \vert j \in J\}\), where \(J\) is a countable index set, be any countable basis.
Let \(t \in T\) be any.
Let us take \(B_t := \{U_j \in B \vert t \in U_j\}\).
\(B_t\) is a set of some open neighborhoods of \(t\).
For each neighborhood of \(t\), \(N'_t \subseteq T\), there is a \(U_j \in B\) such that \(t \in U_j \subseteq N'_t\), because \(B\) is a basis, but \(U_j \in B_t\).
So, \(B_t\) is a neighborhoods basis at \(t\).
\(B_t\) is countable, because it is a subset of countable \(B\).
So, each \(t \in T\) has a countable neighborhoods basis at \(t\).
So, \(T\) is 1st-countable.
