description/proof of that for metric spaces map, map is continuous at point iff map is continuous at point between induced topological spaces
Topics
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of metric spaces map continuous at point.
- The reader knows a definition of topology induced by metric.
- The reader knows a definition of topological spaces map continuous at point.
Target Context
- The reader will have a description and a proof of the proposition that for any metric spaces map, the map is continuous at any point if and only if the map is continuous at the point between the induced topological spaces.
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:
\(M_1\): \(\in \{\text{ the metric spaces }\}\)
\(M_2\): \(\in \{\text{ the metric spaces }\}\)
\(T_1\): \(= \text{ the topological space induced by } M_1\)
\(T_2\): \(= \text{ the topological space induced by } M_2\)
\(m\): \(\in M_1\)
\(f\): \(: M_1 \to M_2\)
\(f'\): \(: T_1 \to T_2, t \mapsto f (t)\)
//
Statements:
\(f \in \{\text{ the maps continuous at } m\}\)
\(\iff\)
\(f' \in \{\text{ the maps continuous at } m\}\)
//
2: Proof
Whole Strategy: Step 1: suppose that \(f\) is continuous at \(m\); Step 2: see that \(f'\) is continuous at \(m\); Step 3: suppose that \(f'\) is continuous at \(m\); Step 4: see that \(f\) is continuous at \(m\).
Step 1:
Let us suppose that \(f\) is continuous at \(m\).
Step 2:
Let \(U_{f (m)} \subseteq T_2\) be any open neighborhood of \(f (m)\).
There is an \(\epsilon \in \mathbb{R}\) such that \(0 \lt \epsilon\) and \(B_{f (m), \epsilon} \subseteq U_{f (m)}\), by the definition of topology induced by metric.
There is a \(\delta \in \mathbb{R}\) such that \(0 \lt \delta\) and \(f (B_{m, \delta}) \subseteq B_{f (m), \epsilon}\), because \(f\) is continuous.
So, \(f' (B_{m, \delta}) = f (B_{m, \delta}) \subseteq B_{f (m), \epsilon} \subseteq U_{f (m)}\).
But \(B_{m, \delta} \subseteq T_1\) is an open neighborhood of \(m\), by Note for the definition of topology induced by metric.
So, \(f'\) is continuous at \(m\).
Step 3:
Let us suppose that \(f'\) is continuous at \(m\).
Step 4:
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
\(B_{f (m), \epsilon} \subseteq T_2\) is open, by Note for the definition of topology induced by metric.
There is an open neighborhood of \(m\), \(U_m \subseteq T_1\), such that \(f' (U_m) \subseteq B_{f (m), \epsilon}\), because \(f'\) is continuous.
There is a \(\delta \in \mathbb{R}\) such that \(0 \lt \delta\) and \(B_{m, \delta} \subseteq U_m\), by the definition of topology induced by metric.
So, \(f (B_{m, \delta}) \subseteq f (U_m) = f' (U_m) \subseteq B_{f (m), \epsilon}\).
So, \(f\) is continuous at \(m\).
