description/proof of that for continuous map between metric spaces, restriction of map on compact domain is uniformly continuous
Topics
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of continuous, metric spaces map.
- 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 metric subspace.
- The reader knows a definition of uniformly continuous map between metric spaces.
Target Context
- The reader will have a description and a proof of the proposition that for any continuous map between any metric spaces with the domain with the induced topology, the restriction of the map on any compact domain is uniformly continuous.
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:
\(M_1\): \(\in \{\text{ the metric spaces }\}\), with the induced topology
\(M_2\): \(\in \{\text{ the metric spaces }\}\)
\(f\): \(: M_1 \to M_2\), \(\in \{\text{ the continuous maps }\}\)
\(K\): \(\in \{\text{ the compact subsets of } M_1\}\), as the metric subspace
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Statements:
\(f \vert_K: K \to M_2 \in \{\text{ the uniformly continuous maps }\}\)
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2: Proof
Whole Strategy: Step 1: take any \(\epsilon\) and for each \(k \in K\), \(\delta (k)\) such that \(f (B_{k, \delta (k)}) \subseteq B_{f (k), \epsilon / 2}\), take any finite subcover of \(K\), \(\{B_{k_j, \delta (k_j) / 2}\}\), and take \(\delta := Min (\{\delta (k_j) / 2\})\); Step 2: see that for each \(k \in K\), \(f (B_{k, \delta}) \subseteq B_{f (k), \epsilon}\); Step 3: conclude that proposition.
Step 1:
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
For each \(k \in K\), there is a \(\delta (k) \in \mathbb{R}\) such that \(0 \lt \delta (k)\) and \(f (B_{k, \delta (k)}) \subseteq B_{f (k), \epsilon / 2}\), because \(f\) is continuous.
\(\{B_{k, \delta (k) / 2} \vert k \in K\}\) is a open cover of \(K\).
As \(K\) is a compact subset, there is a finite subcover, \(\{B_{k_j, \delta (k_j) / 2} \vert j \in J, k_j \in K\}\), where \(J\) is a finite index set.
Let \(\delta := Min (\{\delta (k_j) / 2 \vert j \in J\})\).
Step 2:
Let \(k \in K\) be any.
Let us see that \(f (B_{k, \delta}) \subseteq B_{f (k), \epsilon}\).
\(k \in B_{k_j, \delta (k_j) / 2}\) for a \(j \in J\).
As \(f (B_{k_j, \delta (k_j)}) \subseteq B_{f (k_j), \epsilon / 2}\), \(dist (f (k_j), f (k)) \lt \epsilon / 2\).
Let \(k' \in B_{k, \delta}\) be any.
\(dist (k', k_j) \le dist (k', k) + dist (k, k_j) \lt \delta + \delta (k_j) / 2 \le \delta (k_j) / 2 + \delta (k_j) / 2 = \delta (k_j)\), which means that \(k' \in B_{k_j, \delta (k_j)}\).
So, as \(f (B_{k_j, \delta (k_j)}) \subseteq B_{f (k_j), \epsilon / 2}\), \(dist (f (k'), f (k_j)) \lt \epsilon / 2\).
\(dist (f (k'), f (k)) \le dist (f (k'), f (k_j)) + dist (f (k_j), f (k)) \lt \epsilon / 2 + \epsilon / 2 = \epsilon\).
That means that \(f (B_{k, \delta}) \subseteq B_{f (k), \epsilon}\).
Step 3:
So, \(f \vert_K (B_{k, \delta} \cap K) \subseteq B_{f (k), \epsilon}\).
But \(B_{k, \delta} \cap K\) is \(B_{k, \delta}\) on \(K\), by the subspace metric.
So, \(f \vert_K (B_{k, \delta}) \subseteq B_{f (k), \epsilon}\), where \(B_{k, \delta}\) here is the open ball on \(K\).
So, \(f \vert_K\) is uniformly continuous.
