description/proof of that for metric space, open subset, compact subset contained in open subset, there is positive radius of which open or close ball around each point on compact subset is contained in open subset
Topics
About: metric space
About: topological 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 compact 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 proposition that the compactness of any topological subset as a subset equals the compactness as a subspace.
- The reader admits the proposition that for any metric space and any subset, the closure of the subset is compact if and only if each sequence into the subset has a subsequence that converges in the closure of the subset if and only if the closure of the subset is complete and for each positive real number, there is a set of some finite open balls of the-number-radius that covers the closure of the subset.
Target Context
- The reader will have a description and a proof of the proposition that for any metric space, any open subset, and any compact subset contained in the open subset, there is a positive radius of which the open or closed ball around each point on the compact subset is contained in the open 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
\(U\): \(\in \{\text{ the open subsets of } M\}\)
\(K\): \(\in \{\text{ the compact subsets of } M\}\), such that \(K \subseteq U\)
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Statements:
\(\exists \delta_0 \in \mathbb{R} \text{ such that } 0 \lt \delta_0 (\forall k \in K (B_{k, \delta_0} \subseteq U))\)
\(\land\)
\(\exists \delta'_0 \in \mathbb{R} \text{ such that } 0 \lt \delta'_0 (\forall k \in K (B'_{k, \delta'_0} \subseteq U))\), where \(B'_{k, \delta'_0}\) is the closed ball
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2: Proof
Whole Strategy: apply the proposition that for any metric space and any subset, the closure of the subset is compact if and only if each sequence into the subset has a subsequence that converges in the closure of the subset if and only if the closure of the subset is complete and for each positive real number, there is a set of some finite open balls of the-number-radius that covers the closure of the subset; Step 1: suppose that there was no such \(\delta_0\), and find a contradiction; Step 2: take \(\delta'_0\) such that \(\delta'_0 \lt \delta_0\).
Step 1:
Let us suppose that there was no such \(\delta_0\).
For each \(\delta \in \mathbb{R}\) such that \(0 \lt \delta\), there would be a \(k_\delta \in K\) such that \(\lnot B_{k_\delta, \delta} \subseteq U\).
Let us take for each \(j \in \mathbb{N}\), \(\delta_j := 2^{- j}\).
\(s: \mathbb{N} \to K, j \mapsto k_{\delta_j}\) would be a sequence into \(K\).
\(K \subseteq M\) would be a compact subspace, by the proposition that the compactness of any topological subset as a subset equals the compactness as a subspace.
There would be a subsequence, \(s \circ f\), where \(f: \mathbb{N} \to \mathbb{N}\), that converges to a \(k \in K\), by the immediate corollary mentioned in the proposition that for any metric space and any subset, the closure of the subset is compact if and only if each sequence into the subset has a subsequence that converges in the closure of the subset if and only if the closure of the subset is complete and for each positive real number, there is a set of some finite open balls of the-number-radius that covers the closure of the subset.
There would be an \(\epsilon \in \mathbb{R}\) such that \(0 \lt \epsilon\) and \(B_{k, \epsilon} \subseteq U\), because \(U\) was open.
There would be an \(N \in \mathbb{N}\) such that for each \(j \in \mathbb{N}\) such that \(N \lt j\), \(s \circ f (j) \in B_{k, \epsilon / 2}\), because \(s \circ f\) converged to \(k\).
There would be an \(N' \in \mathbb{N}\) such that \(N \le N'\) and for each \(j \in \mathbb{N}\) such that \(N' \lt j\), \(\delta_{f (j)} \lt \epsilon / 2\), because \(\delta_{f (j)} = 2^{- f (j)}\).
For each \(j \in \mathbb{N}\), there would be a point, \(m_j \in B_{k_{\delta_{f (j)}}, \delta_{f (j)}} \setminus U\), because \(B_{k_{\delta_{f (j)}}, \delta_{f (j)}}\) was not contained in \(U\).
For each \(j \in \mathbb{N}\) such that \(N' \lt j\), \(dist (k, m_j) \le dist (k, k_{\delta_{f (j)}}) + dist (k_{\delta_{f (j)}}, m_j) \lt \epsilon / 2 + \delta_{f (j)} \lt \epsilon / 2 + \epsilon / 2 = \epsilon\), which would mean that \(m_j \in B_{k, \epsilon} \subseteq U\), a contradiction against \(m_j \in B_{k_{\delta_{f (j)}}, \delta_{f (j)}} \setminus U\).
So, the supposition was wrong, and there is a \(\delta_0\).
Step 2:
There is a \(\delta_0\), by Step 1.
Let \(\delta'_0 \in \mathbb{R}\) be any such that \(0 \lt \delta'_0\) and \(\delta'_0 \lt \delta_0\).
Then, for each \(k \in K\), \(B'_{k, \delta'_0} \subseteq B_{k, \delta_0}\).
So, \(B'_{k, \delta'_0} \subseteq B_{k, \delta_0} \subseteq U\).
