description/proof of that for metric space with induced topology and subset, closure of subset is set of convergences of convergent points sequences on subset
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 closure of subset of topological space.
- The reader admits the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.
Target Context
- The reader will have a description and a proof of the proposition that for any metric space with the induced topology and any subset, the closure of the subset is the set of the convergences of the convergent points sequences on the subset.
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 topology induced by the metric
\(S\): \(\subseteq M\)
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Statements:
\(\overline{S} = \{\text{ the convergences of the convergent points sequences on } S\}\)
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2: Proof
Whole Strategy: Step 1: for each \(m \in \overline{S}\), take a convergent points sequence on \(S\) whose convergence is \(m\); Step 2: see that for each convergent points sequence on \(S\) whose convergence is \(m\), \(m \in \overline{S}\); Step 3: conclude the proposition.
Step 1:
Let \(m \in \overline{S}\) be any.
Let us take any points sequence, \(s: \mathbb{N} \to S\), such that for each \(n \in \mathbb{N}\), \(s (n) \in B_{m, 1 / (n + 1)}\), which is possible, because \(B_{m, 1 / (n + 1)} \cap S \neq \emptyset\), by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.
\(s\) is a convergent points sequence whose convergence is \(m\), because for each \(B_{m, \epsilon}\), there is an \(N \in \mathbb{N}\) such that \(1 / (N + 1) \lt \epsilon\), and for each \(n \in \mathbb{N}\) such that \(N \lt n\), \(1 / (n + 1) \lt 1 / (N + 1) \lt \epsilon\), so, \(s (n) \in B_{m, 1 / (n + 1)} \subseteq B_{m, \epsilon}\).
So, \(m \in \{\text{ the convergences of the convergent points sequences on } S\}\).
So, \(\overline{S} \subseteq \{\text{ the convergences of the convergent points sequences on } S\}\).
Step 2:
Let \(m \in \{\text{ the convergences of the convergent points sequences on } S\}\) be any.
\(m\) is the convergence of a convergent points sequence, \(s: \mathbb{N} \to S\).
When \(m \in S\), \(m \in \overline{S}\).
When \(m \notin S\), for each open neighborhood of \(m\), \(U_m \subseteq M\), there is a \(B_{m, \epsilon} \subseteq M\) such that \(B_{m, \epsilon} \subseteq U_m\), and there is an \(n \in \mathbb{N}\) such that \(s (n) \in B_{m, \epsilon}\), because \(s\) converges to \(m\), so, \(U_m \cap S \neq \emptyset\), because \(s (n) \in B_{m, \epsilon} \cap S \subseteq U_m \cap S\), which means that \(m\) is an accumulation point of \(S\), so, \(m \in \overline{S}\), by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.
So, anyway, \(m \in \overline{S}\).
So, \(\{\text{ the convergences of the convergent points sequences on } S\} \subseteq \overline{S}\).
Step 3:
So, \(\overline{S} = \{\text{ the convergences of the convergent points sequences on } S\}\).
