description/proof of that for metric space and subset, distance between subset and point satisfies triangle inequality w.r.t. other point
Topics
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of distance between subset and point on metric space.
- The reader admits the proposition that for any partially-ordered ring, any subset, and any element, if the supremum of the subset exists, the supremum of (the subset plus the element) exists and equals (the supremum of the subset) plus the element, and if the infimum of the subset exists, the infimum of (the subset plus the element) exists and equals (the infimum of the subset) plus the element.
Target Context
- The reader will have a description and a proof of the proposition that for any metric space and any subset, the distance between the subset and any point satisfies the triangle inequality with respect to any other point.
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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Statements:
\(\forall m, m' \in M (dist (S, m) \le dist (S, m') + dist (m', m))\)
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2: Note
Compare with the proposition that for any metric space, a distance between some subsets does not necessarily satisfy the triangle inequality.
3: Proof
Whole Strategy: Step 1: take the definition, \(dist (S, m) = Inf (\{dist (s, m) \vert s \in S\})\), and see that \(Inf (\{dist (s, m) \vert s \in S\}) \le Inf (\{dist (s, m') + dist (m', m) \vert s \in S\}) = dist (m', m) + Inf (\{dist (s, m') \vert s \in S\})\).
Step 1:
By definition, \(dist (S, m) = dist (S, \{m\}) = Inf (\{dist (s, m) \vert s \in S, m \in \{m\}\}) = Inf (\{dist (s, m) \vert s \in S\})\).
As \(dist (s, m) \le dist (s, m') + dist (m', m)\), \(Inf (\{dist (s, m) \vert s \in S\}) \le Inf (\{dist (s, m') + dist (m', m) \vert s \in S\}) = dist (m', m) + Inf (\{dist (s, m') \vert s \in S\})\), by the proposition that for any partially-ordered ring, any subset, and any element, if the supremum of the subset exists, the supremum of (the subset plus the element) exists and equals (the supremum of the subset) plus the element, and if the infimum of the subset exists, the infimum of (the subset plus the element) exists and equals (the infimum of the subset) plus the element, \(= dist (m', m) + dist (S, m')\).
So, \(dist (S, m) \le dist (S, m') + dist (m', m)\).
