description/proof of that for metric space with induced topology, closure of bounded subset is bounded with same diameter
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 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.
- The reader admits the proposition that for any partially-ordered set, any subset, and any subset of the subset, if the infimum of the subset and the infimum of the subset of the subset exist, the infimum of the subset is equal to or smaller than the infimum of the subset of the subset, and if the supremum of the subset and the supremum of the subset of the subset exist, the supremum of the subset is equal to or larger than the supremum of the subset 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, the closure of any bounded subset is bounded with the diameter of 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 induced topology
\(S\): \(\in \{\text{ the bounded subsets of } M\}\), with diameter, \(D\)
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Statements:
\(\overline{S} \in \{\text{ the bounded subsets of } M\}\), with diameter, \(D\)
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2: Proof
Whole Strategy: Step 1: take any \(s_1, s_2 \in \overline{S}\), and see that \(dist (s_1, s_2) \lt D + 2 \epsilon\); Step 2: see that the diameter of \(\overline{S}\) is \(D\).
Step 1:
Let \(s_1, s_2 \in \overline{S}\) be any.
\(s_j\) is in \(S\) or is an accumulation point of \(S\), by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
When \(s_j \in S\), let \(s'_j := s_j\).
Otherwise, there is an \(s'_j \in B_{s_j, \epsilon} \cap S\), because \(s_j\) is an accumulation point.
So, anyway, \(s'_j \in S\).
\(dist (s_1, s_2) \le dist (s_1, s'_1) + dist (s'_1 , s'_2) + dist (s'_2, s_2) \lt \epsilon + D + \epsilon = D + 2 \epsilon\).
As \(\epsilon\) is chosen independent of \(s_1, s_2\), \(r' := D + 2 \epsilon\) will do for \(\overline{S}\), so, \(\overline{S}\) is bounded.
Step 2:
Let us see that the diameter of \(\overline{S}\) is \(D\).
Let the diameter of \(\overline{S}\) be \(D'\).
\(D = Inf (\{r \in \mathbb{R} \vert \forall s_1, s_2 \in S (dist (s_1, s_2) \lt r)\})\).
\(D' = Inf (\{r' \in \mathbb{R} \vert \forall s_1, s_2 \in \overline{S} (dist (s_1, s_2) \lt r')\})\).
For each \(r\) for \(D\), \(D' \le r\), because if \(r \lt D'\), there would be a positive \(\epsilon\) such that \(r + 2 \epsilon \lt D'\), but \(r + 2 \epsilon\) would be an \(r'\), because \(D \le r\) and \(D + 2 \epsilon \le r + 2 \epsilon\), so, \(r' \lt D'\), a contradiction against that \(D'\) was the infimum.
That means that \(D'\) is a lower bound of \(\{r \in \mathbb{R} \vert \forall s_1, s_2 \in S (dist (s_1, s_2) \lt r)\}\), so, \(D' \le D\), because \(D\) is the maximum of such lower bounds.
\(\{r' \in \mathbb{R} \vert \forall s_1, s_2 \in \overline{S} (dist (s_1, s_2) \lt r')\} \subseteq \{r \in \mathbb{R} \vert \forall s_1, s_2 \in S (dist (s_1, s_2) \lt r)\}\), because \(dist (s_1, s_2) \lt r'\) for \(\overline{S}\) implies that \(dist (s_1, s_2) \lt r'\) for \(S\), because \(S \subseteq \overline{S}\).
So, \(D \le D'\), by the proposition that for any partially-ordered set, any subset, and any subset of the subset, if the infimum of the subset and the infimum of the subset of the subset exist, the infimum of the subset is equal to or smaller than the infimum of the subset of the subset, and if the supremum of the subset and the supremum of the subset of the subset exist, the supremum of the subset is equal to or larger than the supremum of the subset of the subset.
So, \(D' = D\).
