definition of totally bounded subset of metric space
Topics
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of open ball around point on metric space.
Target Context
- The reader will have a definition of totally bounded subset of metric 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:
\( M\): \(\in \{\text{ the metric spaces }\}\)
\(*S\): \(\subseteq M\)
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Conditions:
\(\forall \epsilon \in \mathbb{R} \text{ such that } 0 \lt \epsilon (\exists J \in \{\text{ the finite index sets }\}, \exists \{B_{s_j, \epsilon} \vert j \in J, s_j \in S\} (S \subseteq \cup_{j \in J} B_{s_j, \epsilon}))\)
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2: Note
Let us see that any totally bounded \(S\) is bounded.
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
There is a \(\{B_{s_j, \epsilon} \vert j \in J, s_j \in S\}\).
Let \(d := Max (\{dist (s_j, s_l) \vert j, l \in J\})\), which exists in \(\mathbb{R}\), because \(J\) is finite.
Let \(s, s' \in S\) be any.
\(s \in B_{s_j, \epsilon}\) and \(s' \in B_{s_l, \epsilon}\).
\(dist (s, s') \le dist (s, s_j) + dist (s_j, s_l) + dist (s_l, s') \lt \epsilon + d + \epsilon = d + 2 \epsilon\).
As \(d\) and \(\epsilon\) do not depend on \(s, s'\), \(S\) is bounded.
A bounded subset of a metric space is not necessarily totally bounded.
For example, let \(M\) be any infinite set with the metric such that for each \(m_1, m_2 \in M\), when \(m_1 = m_2\), \(dist (m_1, m_2) = 0\), and otherwise, \(dist (m_1, m_2) = 1\).
\(dist\) is indeed a metric: for each \(m_1, m_2, m_3 \in M\), 1) \(0 \le dist (m_1, m_2)\) and \(dist (m_1, m_2) = 0 \iff m_1 = m_2\); 2) \(dist (m_1, m_2) = dist (m_2, m_1)\); 3) \(dist (m_1, m_3) \le dist (m_1, m_2) + dist (m_2, m_3)\): when \(m_1 = m_3\), \(dist (m_1, m_3) = 0 \le dist (m_1, m_2) + dist (m_2, m_3)\), and otherwise, \(dist (m_1, m_3) = 1\) and \(1 \le dist (m_1, m_2) + dist (m_2, m_3)\), because \(m_1 \neq m_2\) or \(m_2 \neq m_3\).
Then, \(M\) is bounded, because for each \(m_1, m_2 \in M\), \(dist (m_1, m_2) \lt 2\).
But \(M\) is not totally bounded, because for any \(\epsilon \in \mathbb{R}\) such that \(0 \lt \epsilon \lt 1\), for each \(m \in M\), \(B_{m, \epsilon} = \{m\}\), so, all the \(B_{m, \epsilon}\) s are necessary to cover \(M\).
