description/proof of that for metric space, closed ball is closed and contains but not necessarily equals closure of open ball with same radius
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 knows a definition of closed ball around point on metric space.
- The reader admits the local criterion for openness.
Target Context
- The reader will have a description and a proof of the proposition that for any metric space with the induced topology, any closed ball is closed and contains but not necessarily equal the closure of the open ball with the same radius.
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
\(m\): \(\in M\)
\(\epsilon\): \(\in \mathbb{R}\), such that \(0 \lt \epsilon\)
\(B'_{m, \epsilon}\): \(\in \{\text{ the closed balls at } m\}\)
\(B_{m, \epsilon}\): \(\in \{\text{ the open balls at } m\}\)
//
Statements:
\(B'_{m, \epsilon} \in \{\text{ the closed subsets of } M\}\)
\(\land\)
\(\overline{B_{m, \epsilon}} \subseteq B'_{m, \epsilon}\)
//
2: Note
Of course, on any Euclidean metric space, \(\overline{B_{m, \epsilon}} = B'_{m, \epsilon}\), but we need to be aware that that is not necessarily the case on a general metric space.
3: Proof
Whole Strategy: Step 1: see that \(B'_{m, \epsilon}\) is closed and contains \(B_{m, \epsilon}\); Step 2: see an example that \(B'_{m, \epsilon}\) does not equal \(\overline{B_{m, \epsilon}}\).
Step 1:
Let us see that \(B'_{m, \epsilon} \subseteq M\) is closed.
Let \(p \in M \setminus B'_{m, \epsilon}\) be any.
Let \(r := dist (m, p)\).
\(\epsilon \lt r\).
Let us take \(B_{p, (r - \epsilon) / 2} \subseteq M\).
\(B_{p, (r - \epsilon) / 2} \subseteq M \setminus B'_{m, \epsilon}\), because for each \(p' \in B_{p, (r - \epsilon) / 2}\), \(r = dist (m, p) \le dist (m, p') + dist (p', p) \lt dist (m, p') + (r - \epsilon) / 2\), so, \(r - (r - \epsilon) / 2 \lt dist (m, p')\), but the left hand side is \((r + \epsilon) / 2\) and \(\epsilon \lt (r + \epsilon) / 2\), so, \(\epsilon \lt dist (m, p')\), which means that \(p' \notin B'_{m, \epsilon}\).
So, by the local criterion for openness, \(M \setminus B'_{m, \epsilon} \subseteq M\) is open, so, \(B'_{m, \epsilon} \subseteq M\) is closed.
Obviously, \(B_{m, \epsilon} \subseteq B'_{m, \epsilon}\).
As \(\overline{B_{m, \epsilon}}\) is the intersection of all the closed subsets that contain \(B_{m, \epsilon}\), \(\overline{B_{m, \epsilon}} \subseteq B'_{m, \epsilon}\).
Step 2:
Let us see an example that \(B'_{m, \epsilon}\) does not equal \(\overline{B_{m, \epsilon}}\).
Let \(\mathbb{R}^2\) be the Euclidean metric space.
Let \(M \subset \mathbb{R}^2\) be the metric subspace, \(\{p \in \mathbb{R}^2 \vert dist (0, p) \lt 1\} \cup \{p \in \mathbb{R}^2 \vert 2 \le dist (0, p)\}\).
Note that \(\{p \in \mathbb{R}^2 \vert dist (0, p) \lt 1\} = B_{0, 1}\).
Then, \(B_{0, 2}\) is \(B_{0, 1}\), which is closed, because for each \(p' \in M \setminus B_{0, 1}\), \(B_{p', 1 / 2} \subseteq M \setminus B_{0, 1}\), so, \(M \setminus B_{0, 1}\) is open.
So, \(\overline{B_{0, 2}} = B_{0, 2}\).
But \(B'_{0, 2}\) does not equal \(B_{0, 2}\).
