description/proof of that for continuous map from topological space into \(1\)-dimensional Euclidean topological space, its absolute map is continuous
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of continuous, topological spaces map.
- The reader knows a definition of Euclidean topological space.
Target Context
- The reader will have a description and a proof of the proposition that for any continuous map from any topological space into the \(1\)-dimensional Euclidean topological space, its absolute map 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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(\mathbb{R}\): \(= \text{ the Euclidean topological space }\)
\(f\): \(: T \to \mathbb{R}\), \(\in \{\text{ the continuous maps }\}\)
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Statements:
\(\vert f \vert: T \to \mathbb{R}, t \mapsto \vert f (t) \vert \in \{\text{ the continuous maps }\}\)
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2: Proof
Whole Strategy: Step 1: let \(t \in T\) be any and let \(N_{\vert f \vert (t)} \subseteq \mathbb{R}\) be any neighborhood of \(\vert f \vert (t)\); Step 2: suppose that \(0 \lt \vert f (t) \vert\), and take any open ball around \(\vert f \vert (t)\), \(B_{\vert f \vert (t), \epsilon}\), such that \(\epsilon \lt \vert f (t) \vert\) and \(B_{\vert f \vert (t), \epsilon} \subseteq N_{\vert f \vert (t)}\); Step 3: take an open neighborhood of \(t\), \(U_t\), such that \(f (U_t) \subseteq B_{f (t), \epsilon}\), and see that \(\vert f \vert (U_t) \subseteq B_{\vert f \vert (t), \epsilon}\); Step 4: suppose that \(0 = \vert f (t) \vert\), and take any open ball around \(\vert f \vert (t)\), \(B_{\vert f \vert (t), \epsilon}\), such that \(B_{\vert f \vert (t), \epsilon} \subseteq N_{\vert f \vert (t)}\); Step 5: take an open neighborhood of \(t\), \(U_t\), such that \(f (U_t) \subseteq B_{f (t), \epsilon}\), and see that \(\vert f \vert (U_t) \subseteq B_{\vert f \vert (t), \epsilon}\); Step 6: conclude the proposition.
Step 1:
Let \(t \in T\) be any.
Let \(N_{\vert f \vert (t)} \subseteq \mathbb{R}\) be any neighborhood of \(\vert f \vert (t)\).
Step 2:
Let us suppose that \(0 \lt \vert f (t) \vert\).
Let us take any open ball around \(\vert f \vert (t)\), \(B_{\vert f \vert (t), \epsilon}\), such that \(\epsilon \lt \vert f (t) \vert\) and \(B_{\vert f \vert (t), \epsilon} \subseteq N_{\vert f \vert (t)}\).
Step 3:
As \(f\) is continuous, there is an open neighborhood of \(t\), \(U_t \subseteq T\), such that \(f (U_t) \subseteq B_{f (t), \epsilon}\).
Note that \(B_{f (t), \epsilon}\) is all positive or all negative and so is \(f (U_t)\).
Let us see that \(\vert f \vert (U_t) \subseteq B_{\vert f \vert (t), \epsilon}\).
Let \(u \in U_t\) be any.
Let us think of \(dist (\vert f \vert (u), \vert f \vert (t))\).
When \(0 \lt f (t)\), \(0 \lt f (u)\), and \(dist (\vert f \vert (u), \vert f \vert (t)) = dist (f (u), f (t)) \lt \epsilon\).
When \(f (t) \lt 0\), \(f (u) \lt 0\), and \(dist (\vert f \vert (u), \vert f \vert (t)) = dist (- f (u), - f (t)) = \vert - f (u) - (- f (t)) \vert = \vert - f (u) + f (t) \vert = \vert f (u) - f (t) \vert = dist (f (u), f (t)) \lt \epsilon\).
That means that \(\vert f \vert (U_t) \subseteq B_{\vert f \vert (t), \epsilon}\).
Step 4:
Let us suppose that \(0 = \vert f (t) \vert\).
Let us take any open ball around \(\vert f \vert (t)\), \(B_{\vert f \vert (t), \epsilon}\), such that \(B_{\vert f \vert (t), \epsilon} \subseteq N_{\vert f \vert (t)}\).
Step 5:
As \(f\) is continuous, there is an open neighborhood of \(t\), \(U_t\), such that \(f (U_t) \subseteq B_{f (t), \epsilon}\).
Note that in this case, \(B_{f (t), \epsilon} = B_{0, \epsilon} = B_{\vert f \vert (t), \epsilon}\).
Let us see that \(\vert f \vert (U_t) \subseteq B_{\vert f \vert (t), \epsilon}\).
Let \(u \in U_t\) be any.
\(dist (\vert f \vert (u), \vert f \vert (t)) = dist (\vert f \vert (u), 0)\), which equals \(dist (f (u), 0)\) or \(dist (- f (u), 0)\).
\(dist (f (u), 0) \lt \epsilon\).
\(dist (- f (u), 0) = dist (f (u), 0) \lt \epsilon\).
So, \(dist (\vert f \vert (u), \vert f \vert (t)) \lt \epsilon\).
That means that \(\vert f \vert (U_t) \subseteq B_{\vert f \vert (t), \epsilon}\).
Step 6:
So, in any case, \(\vert f \vert (U_t) \subseteq B_{\vert f \vert (t), \epsilon} \subseteq N_{\vert f \vert}\).
So, \(\vert f \vert\) is continuous.
