description/proof of that for metric space and subset, subset is totally bounded iff closure of subset is totally bounded iff closure of subset is totally bounded with centers in 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 knows a definition of totally bounded subset of metric 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 and any subset, the subset is totally bounded if and only if the closure of the subset is totally bounded if and only if the closure of the subset is totally bounded with the centers in 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
\(S\): \(\subseteq M\)
\(\overline{S}\): \(= \text{ the closure }\)
//
Statements:
\(S \in \{\text{ the totally bounded subsets of } M\}\)
\(\iff\)
\(\overline{S} \in \{\text{ the totally bounded subsets of } M\}\)
\(\iff\)
\(\overline{S} \in \{\text{ the totally bounded subsets of } M\}\) with the centers chosen in \(S\)
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2: Proof
Whole Strategy: Step 1: suppose \(S\) is totally bounded; Step 2: see that \(\overline{S}\) is totally bounded with the centers chosen in \(S\), by taking any finite cover of \(S\) for \(\epsilon / 2\); Step 3: suppose that \(\overline{S}\) is totally bounded; Step 4: see that \(S\) is totally bounded, by taking any finite cover of \(\overline{S}\) for \(\epsilon / 2\) and for each center, \(m_j\), choosing any \(s_j \in B_{m_j, \epsilon / 2} \cap S\); Step 5: conclude the proposition.
Step 1:
Let us suppose \(S\) is totally bounded.
Step 2:
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
There are a finite index set, \(J\), and some \(\{B_{s_j, \epsilon / 2} \vert j \in J, s_j \in S\}\) such that \(S \subseteq \cup_{j \in J} B_{s_j, \epsilon / 2}\).
Let \(m \in \overline{S}\) be any.
\(B_{m, \epsilon / 2} \cap S \neq \emptyset\), by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset, so, there is an \(s \in B_{m, \epsilon / 2} \cap S\).
\(s \in B_{s_j, \epsilon / 2}\) for a \(j \in J\).
\(dist (m, s_j) \le dist (m, s) + dist (s, s_j) \lt \epsilon / 2 + \epsilon / 2 = \epsilon\).
That means that \(\{B_{s_j, \epsilon} \vert j \in J\}\) covers \(\overline{S}\).
So, \(\overline{S}\) is totally bounded with the centers chosen in \(S\).
Step 3:
Let us suppose that \(\overline{S}\) is totally bounded.
Step 4:
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
There are a finite index set, \(J\), and some \(\{B_{m_j, \epsilon / 2} \vert j \in J, m_j \in \overline{S}\}\) such that \(\overline{S} \subseteq \cup_{j \in J} B_{m_j, \epsilon / 2}\).
For each \(j \in J\), \(B_{m_j, \epsilon / 2} \cap S \neq \emptyset\), by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset, so, there is an \(s_j \in B_{m_j, \epsilon / 2} \cap S\) for each \(j \in J\).
Let \(s \in S\) be any.
As \(s \in \overline{S}\), \(s \in B_{m_j, \epsilon / 2}\) for a \(j \in J\), and with the \(s_j\) chosen above, \(dist (s, s_j) \le dist (s, m_j) + dist (m_j, s_j) \lt \epsilon / 2 + \epsilon / 2 = \epsilon\), so, \(s \in B_{s_j, \epsilon}\).
That means that \(\{B_{s_j, \epsilon} \vert j \in J, s_j \in S\}\) covers \(S\).
So, \(S\) is totally bounded.
Step 5:
If \(\overline{S}\) is totally bounded with the centers chosen in \(S\), \(\overline{S}\) is totally bounded, obviously.
If \(S\) is totally bounded, \(\overline{S}\) is totally bounded with the centers chosen in \(S\), by Step 2, and \(\overline{S}\) is totally bounded.
If \(\overline{S}\) is totally bounded, \(S\) is totally bounded, by Step 4.
If \(\overline{S}\) is totally bounded, \(S\) is totally bounded, by Step 4, and \(\overline{S}\) is totally bounded with the centers chosen in \(S\), by Step 2.
If \(\overline{S}\) is totally bounded with the centers chosen in \(S\), \(\overline{S}\) is totally bounded, as has been mentioned at the top of Step 5.
So, the proposition holds.
