description/proof of that for metric space with induced topology, points sequence converges to point as on metric space iff it converges to point as on topological space
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 convergence of sequence on metric space.
- The reader knows a definition of convergence of net with directed index set.
Target Context
- The reader will have a description and a proof of the proposition that for any metric space with the induced topology, any points sequence converges to any point as on the metric space if and only if the sequence converges to the point as on the topological 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:
\(M\): \(\in \{\text{ the metric spaces }\}\), with the induced topology
\(s\): \(: \mathbb{N} \to M\)
\(m\): \(\in M\)
//
Statements:
\(s \text{ converges to } m \text{ with } M \text{ as the metric space }\)
\(\iff\)
\(s \text{ converges to } m \text{ with } M \text{ as the topological space }\)
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2: Note
Convergence of points sequence on metric space and convergence of points sequence on topological space (which is a convergence of net with directed index set: any points sequence on any topological space is a net with directed index set) have some independent definitions.
On any metric space with the induced topology, do they really coincide? This proposition says yes.
3: Proof
Whole Strategy: Step 1: suppose that \(s\) converges to \(m\) with \(M\) as the metric space; Step 2: see that \(s\) converges to \(m\) with \(M\) as the topological space; Step 3: suppose that \(s\) converges to \(m\) with \(M\) as the topological space; Step 4: see that \(s\) converges to \(m\) with \(M\) as the metric space.
Step 1:
Let us suppose that \(s\) converges to \(m\) with \(M\) as the metric space.
Step 2:
Let \(N_m \subseteq M\) be any neighborhood of \(m\).
There is an open neighborhood of \(m\), \(U_m \subseteq M\), such that \(U_m \subseteq N_m\).
There is a \(B_{m, \epsilon} \subseteq M\) such that \(B_{m, \epsilon} \subseteq U_m\), because the topology is induced by the metric.
There is an \(N \in \mathbb{N}\) such that for each \(j \in \mathbb{N}\) such that \(N \lt j\), \(s (j) \in B_{m, \epsilon}\).
So, for each \(j \in \mathbb{N}\) such that \(N \lt j\), \(s (j) \in B_{m, \epsilon} \subseteq U_m \subseteq N_m\).
So, \(s\) converges to \(m\) with \(M\) as the topological space.
Step 3:
Let us suppose that \(s\) converges to \(m\) with \(M\) as the topological space.
Step 4:
Let \(B_{m, \epsilon} \subseteq M\) be any such that \(0 \lt \epsilon\).
\(B_{m, \epsilon}\) is a neighborhood of \(m\), because the topology is induced by the metric.
There is an \(N \in \mathbb{N}\) such that for each \(j \in \mathbb{N}\) such that \(N \lt j\), \(s (j) \in B_{m, \epsilon}\).
That means that \(s\) converges to \(m\) with \(M\) as the metric space.
