description/proof of that for metric space and subset, set of points whose distances to subset are \(0\) is closure of subset
Topics
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of metric space.
- The reader knows a definition of topology induced by metric.
- The reader knows a definition of closure of subset of topological 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 set of the points whose distances to the subset are \(0\) is the closure 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:
\(T\): \(\in \{\text{ the metric spaces }\}\), with the induced topology
\(S\): \(\subseteq T\)
\(R\): \(= \{r \in T \vert dist (r, S) = 0\}\)
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Statements:
\(R = \overline{S}\)
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2: Proof
Whole Strategy: Step 1: see that \(R \subseteq \overline{S}\); Step 2: see that \(\overline{S} \subseteq R\).
Step 1:
Let \(r \in R\) be any.
Let \(U_r \subseteq T\) be any open neighborhood of \(r\).
There is an \(\epsilon \in \mathbb{R}\) such that \(0 \lt \epsilon\) and \(B_{r, \epsilon} \subseteq U_r\), where \(B_{r, \epsilon}\) is the \(\epsilon\)-'open ball' around \(r\).
\(dist (r, S) := inf \{dist (r, s) \vert s \in S\}\), so, \(dist (r, S) = 0\) means that there is an \(s \in S\) such that \(dist (r, s) \lt \epsilon\), which means that \(s \in B_{r, \epsilon}\).
So, \(B_{r, \epsilon} \cap S \neq \emptyset\), which implies that \(U_r \cap S \neq \emptyset\).
So, \(r \in \overline{S}\), by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.
So, \(R \subseteq \overline{S}\).
Step 2:
Let \(s' \in \overline{S}\) be any.
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
\(B_{s', \epsilon} \cap S \neq \emptyset\), by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.
That means that there is an \(s \in S\) such that \(s \in B_{s', \epsilon}\), which means that \(dist (s', s) \lt \epsilon\).
That means that \(inf \{dist (s', s) \vert s \in S\} = 0\), which means that \(dist (s', S) = 0\).
So, \(s' \in R\).
So, \(\overline{S} \subseteq R\).
