description/proof of that for metric space, subset, subset of subset, and positive real number, if open ball around each point of subset of subset with number-radius is contained in subset, closure of union of open balls around each point of subset of subset with smaller-radius is contained 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 open ball around point on 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 set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset.
Target Context
- The reader will have a description and a proof of the proposition that for any metric space, any subset, any subset of the subset, and any positive real number, if the open ball around each point of the subset of the subset with the number-radius is contained in the subset, the closure of the union of the open balls around each point of the subset of the subset with any smaller-radius is contained 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
\(S'\): \(\subseteq M\)
\(S\): \(\subseteq S'\)
\(\epsilon\): \(\in \mathbb{R}\), such that \(0 \lt \epsilon\)
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Statements:
\(\forall s \in S (B_{s, \epsilon} \subseteq S')\)
\(\implies\)
\(\forall \delta \in \mathbb{R} \text{ such that } 0 \lt \delta \lt \epsilon (\overline{\cup_{s \in S} B_{s, \delta}} \subseteq S')\)
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2: Proof
Whole Strategy: Step 1: see that for each \(m \in \overline{\cup_{s \in S} B_{s, \delta}}\), \(m\) is in \(\cup_{s \in S} B_{s, \delta}\) or an accumulation point of \(\cup_{s \in S} B_{s, \delta}\) and \(m\) is in \(S'\) anyway.
Step 1:
Let \(m \in \overline{\cup_{s \in S} B_{s, \delta}}\) be any.
\(m\) is in \(\cup_{s \in S} B_{s, \delta}\) or an accumulation point of \(\cup_{s \in S} B_{s, \delta}\), by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.
When \(m\) is in \(\cup_{s \in S} B_{s, \delta}\), \(m \in B_{s, \delta}\) for an \(s \in S\), and \(m \in B_{s, \delta} \subseteq B_{s, \epsilon} \subseteq S'\).
Let us suppose that \(m\) is an accumulation point of \(\cup_{s \in S} B_{s, \delta}\).
There is a \(p \in B_{m, \epsilon - \delta} \cap \cup_{s \in S} B_{s, \delta}\).
\(B_{m, \epsilon - \delta} \cap \cup_{s \in S} B_{s, \delta} = \cup_{s \in S} (B_{m, \epsilon - \delta} \cap B_{s, \delta})\), by the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset.
So, \(p \in B_{m, \epsilon - \delta} \cap B_{s, \delta}\) for an \(s \in S\).
\(dist (m, s) \le dist (m, p) + dist (p, s) \lt \epsilon - \delta + \delta = \epsilon\), so, \(m \in B_{s, \epsilon} \subseteq S'\).
So, \(m \in S'\), anyway.
So, \(\overline{\cup_{s \in S} B_{s, \delta}} \subseteq S'\).
