description/proof of that for uniformly convergent sequence of continuous maps from topological space into metric space, convergence is continuous
Topics
About: topological space
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of uniformly convergent sequence of maps from topological space into metric space.
- The reader knows a definition of topology induced by metric.
- The reader knows a definition of continuous, topological spaces map.
Target Context
- The reader will have a description and a proof of the proposition that for any uniformly convergent sequence of continuous maps from any topological space into any metric space with the induced topology, the convergence is 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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the metric spaces }\}\), with the induced topology
\(J\): \(\subseteq \mathbb{N}\)
\(s\): \(: J \to \{g: T_1 \to T_2\}\), \(\in \{\text{ the uniformly convergent sequences }\}\)
\(f\): \(: T_1 \to T_2\), \(= \text{ the convergence of } s\)
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Statements:
\(\forall j \in J (s (j) \in \{\text{ the continuous maps }\})\)
\(\implies\)
\(f \in \{\text{ the continuous maps }\}\)
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2: Note
In fact, some finite number of \(s (j)\) s do not need to be continuous, because they do not influence \(f\): we can take another sequence that has excluded the non-continuous maps, and the convergence is obviously \(f\).
3: Proof
Whole Strategy: Step 1: see that for each \(t, t' \in T_1\) and each \(j \in J\), \(dist (f (t'), f (t)) \le dist (f (t'), s (j) (t')) + dist (s (j) (t'), s (j) (t)) + dist (s (j) (t), f (t))\); Step 2: take \(N\) such that for each \(N \lt j\), for each \(t'' \in T_1\), \(dist (s (j) (t''), f (t'')) \lt \epsilon / 3\), and for any fixed such \(j\), take an open neighborhood of \(t\), \(U_{j, t}\), such that \(s (j) (U_{j, t}) \subseteq B_{s (j) (t), \epsilon / 3}\).
Step 1:
Let \(t, t' \in T_1\) be any and let \(j \in J\) be any.
\(dist (f (t'), f (t)) \le dist (f (t'), s (j) (t')) + dist (s (j) (t'), s (j) (t)) + dist (s (j) (t), f (t))\).
Step 2:
There is an \(N \in \mathbb{N}\) such that for each \(j \in \mathbb{N}\) such that \(N \lt j\), for each \(t'' \in T_1\), \(dist (s (j) (t''), f (t'')) \lt \epsilon / 3\).
Let us take such any \(j\).
Let \(B_{s (j) (t), \epsilon / 3}\) be the \(\epsilon / 3\)-'open ball' around \(s (j) (t)\).
As \(s (j)\) is continuous, there is an open neighborhood of \(t\), \(U_{j, t} \subseteq T_1\), such that \(s (j) (U_{j, t}) \subseteq B_{s (j) (t), \epsilon / 3}\).
For each \(u \in U_{j, t}\), by Step 1, \(dist (f (u), f (t)) \le dist (f (u), s (j) (u)) + dist (s (j) (u), s (j) (t)) + dist (s (j) (t), f (t)) \lt \epsilon / 3 + \epsilon / 3 + \epsilon / 3 = \epsilon\).
So, \(f (U_{j, t}) \subseteq B_{f (t), \epsilon}\).
For each open neighborhood of \(f (t)\), \(U_{f (t)} \subseteq T_2\), there is an open ball around \(f (t)\), \(B_{f (t), \epsilon} \subseteq T_2\), such that \(B_{f (t), \epsilon} \subseteq U_{f (t)}\), and there is a corresponding \(U_{j, t}\) such that \(f (U_{j, t}) \subseteq B_{f (t), \epsilon} \subseteq U_{f (t)}\).
So, \(f\) is continuous.
