description/proof of that metric space is 1st-countable
Topics
About: metric space
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of topology induced by metric.
- The reader knows a definition of 1st-countable topological space.
Target Context
- The reader will have a description and a proof of the proposition that any metric space with the induced topology 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:
\(M\): \(\in \{\text{ the metric spaces }\}\), with the induced topology
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Statements:
\(M \in \{\text{ the 1st-countable topological spaces }\}\)
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2: Proof
Whole Strategy: Step 1: for each \(m \in M\), take \(B_m := \{B_{m, q} \subseteq M \vert q \in \mathbb{Q} \land 0 \lt q\}\), and see that \(B_m\) is a countable neighborhoods basis at \(m\).
Step 1:
Let \(m \in M\) be any.
Let us take \(B_m := \{B_{m, q} \subseteq M \vert q \in \mathbb{Q} \land 0 \lt q\}\).
\(B_{m, q}\) is an open neighborhood of \(m\), by Note for the definition of topology induced by metric.
For each open neighborhood of \(m\), \(U_m \subseteq M\), there is an open ball around \(m\), \(B_{m, \epsilon} \subseteq M\), such that \(B_{m, \epsilon} \subseteq U_m\), by the definition of topology induced by metric.
Then, there is a \(q \in \mathbb{Q}\) such that \(0 \lt q \lt \epsilon\), and \(m \in B_{m, q} \subseteq B_{m, \epsilon} \subseteq M\).
So, \(B_m\) is a neighborhoods basis at \(m\).
\(B_m\) is countable, because \(\mathbb{Q}\) is countable.
So, \(B_m\) is a countable neighborhoods basis at \(m\).
As there is a countable neighborhoods basis at each point, \(M\) is 1st-countable.
