description/proof of that for finite number of continuous maps from same topological space into \(1\)-dimensional Euclidean topological space, maximum or minimum 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.
- The reader admits the proposition that for any continuous map from any topological space into the \(1\)-dimensional Euclidean topological space, its absolute map is continuous.
- The reader admits the proposition that for any maps between any arbitrary subspaces of any topological spaces continuous at any corresponding points, the composition is continuous at the point.
Target Context
- The reader will have a description and a proof of the proposition that for any finite number of continuous maps from any same topological space into the \(1\)-dimensional Euclidean topological space, the maximum map or the minimum 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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(\mathbb{R}\): \(= \text{ the Euclidean topological space }\)
\(\{f_1, ..., f_n\}\): \(f_j: T \to \mathbb{R} \in \{\text{ the continuous maps }\}\)
\(max (\{f_1, ..., f_n\})\): \(: T \to \mathbb{R}, t \mapsto max \{f_1 (t), ..., f_n (t)\}\)
\(min (\{f_1, ..., f_n\})\): \(: T \to \mathbb{R}, t \mapsto min \{f_1 (t), ..., f_n (t)\}\)
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Statements:
\(max (\{f_1, ..., f_n\}) \in \{\text{ the continuous maps }\}\)
\(\land\)
\(min (\{f_1, ..., f_n\}) \in \{\text{ the continuous maps }\}\)
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2: Proof
Whole Strategy: Step 1: see that \(max (\{f_1, ..., f_{n + 1}\}) = max (\{max (\{f_1, ..., f_n\}), f_{n + 1}\})\); Step 2: see that \(max (\{f_1, f_2\})\) is continuous; Step 3: conclude the proposition for \(max (\{f_1, ..., f_n\})\) by the induction principle; Step 4: see that \(min (\{f_1, ..., f_{n + 1}\}) = min (\{min (\{f_1, ..., f_n\}), f_{n + 1}\})\); Step 5: see that \(min (\{f_1, f_2\})\) is continuous; Step 3: conclude the proposition for \(min (\{f_1, ..., f_n\})\) by the induction principle.
Step 1:
Let us see that \(max (\{f_1, ..., f_{n + 1}\}) = max (\{max (\{f_1, ..., f_n\}), f_{n + 1}\})\).
Let \(t \in T\) be any.
\(max (\{f_1, ..., f_{n + 1}\}) (t) = max (\{f_1 (t), ..., f_{n + 1} (t)\})\).
\(max (\{max (\{f_1, ..., f_n\}), f_{n + 1}\}) (t) = max (\{max (\{f_1, ..., f_n\}) (t), f_{n + 1} (t)\}) = max (\{max (\{f_1 (t), ..., f_n (t)\}), f_{n + 1} (t)\})\).
But \(max (\{f_1 (t), ..., f_{n + 1} (t)\}) = max (\{max (\{f_1 (t), ..., f_n (t)\}), f_{n + 1} (t)\})\), because when \(f_{n + 1} (t)\) is the maximum among \(\{f_1 (t), ..., f_{n + 1} (t)\}\), \(max (\{f_1 (t), ..., f_{n + 1} (t)\}) = f_{n + 1} (t)\), while whatever the maximum among \(\{f_1 (t), ..., f_n (t)\}\) is, \(max (\{max (\{f_1 (t), ..., f_n (t)\}), f_{n + 1} (t)\}) = f_{n + 1} (t)\); when \(f_{j} (t)\) where \(j \lt n + 1\) is the maximum among \(\{f_1 (t), ..., f_{n + 1} (t)\}\), \(max (\{f_1 (t), ..., f_{n + 1} (t)\}) = f_{j} (t)\), while \(max (\{max (\{f_1 (t), ..., f_n (t)\}), f_{n + 1} (t)\}) = max (\{f_j (t), f_{n + 1} (t)\}) = f_j (t)\).
So, \(max (\{f_1, ..., f_{n + 1}\}) = max (\{max (\{f_1, ..., f_n\}), f_{n + 1}\})\).
Step 2:
Let us see that \(max (\{f_1, f_2\})\) is continuous.
Let us see that \(max (\{f_1, f_2\}) = 1 / 2 (f_1 + f_2 + \vert f_1 - f_2 \vert)\).
Let \(t \in T\) be any.
\(f_1 (t) \le f_2 (t)\) or \(f_2 (t) \lt f_1 (t)\).
Let us suppose that \(f_1 (t) \le f_2 (t)\).
\(max (\{f_1, f_2\}) (t) = max (\{f_1 (t), f_2 (t)\}) = f_2 (t)\), while \((1 / 2 (f_1 + f_2 + \vert f_1 - f_2 \vert)) (t) = 1 / 2 (f_1 (t) + f_2 (t) + \vert f_1 (t) - f_2 (t) \vert) = 1 / 2 (f_1 (t) + f_2 (t) + f_2 (t) - f_1 (t)) = 1 / 2 (2 f_2 (t)) = f_2 (t)\).
Let us suppose that \(f_2 (t) \lt f_1 (t)\).
\(max (\{f_1, f_2\}) (t) = max (\{f_1 (t), f_2 (t)\}) = f_1 (t)\), while \((1 / 2 (f_1 + f_2 + \vert f_1 - f_2 \vert)) (t) = 1 / 2 (f_1 (t) + f_2 (t) + \vert f_1 (t) - f_2 (t) \vert) = 1 / 2 (f_1 (t) + f_2 (t) + f_1 (t) - f_2 (t)) = 1 / 2 (2 f_1 (t)) = f_1 (t)\).
So, \(max (\{f_1, f_2\}) = 1 / 2 (f_1 + f_2 + \vert f_1 - f_2 \vert)\).
\(f_1 - f_2\) is continuous, as a well-known fact, and \(\vert f_1 - f_2 \vert\) is continuous, by the proposition that for any continuous map from any topological space into the \(1\)-dimensional Euclidean topological space, its absolute map is continuous.
Then, \(1 / 2 (f_1 + f_2 + \vert f_1 - f_2 \vert)\) is continuous, as a well-known fact.
Step 3:
Let us prove the proposition by the induction principle.
When \(n = 2\), \(max (\{f_1, ..., f_n\})\) is continuous by Step 2.
Let us suppose that for any \(n\), \(max (\{f_1, ..., f_n\})\) is continuous.
By Step 1, \(max (\{f_1, ..., f_{n + 1}\}) = max (\{max (\{f_1, ..., f_n\}), f_{n + 1}\})\), which is continuous by Step 2.
So, \(max (\{f_1, ..., f_n\})\) is continuous for each \(n \in \mathbb{N} \setminus \{0\}\).
Step 4:
Let us see that \(min (\{f_1, ..., f_{n + 1}\}) = min (\{min (\{f_1, ..., f_n\}), f_{n + 1}\})\).
Let \(t \in T\) be any.
\(min (\{f_1, ..., f_{n + 1}\}) (t) = min (\{f_1 (t), ..., f_{n + 1} (t)\})\).
\(min (\{min (\{f_1, ..., f_n\}), f_{n + 1}\}) (t) = min (\{min (\{f_1, ..., f_n\}) (t), f_{n + 1} (t)\}) = min (\{min (\{f_1 (t), ..., f_n (t)\}), f_{n + 1} (t)\})\).
But \(min (\{f_1 (t), ..., f_{n + 1} (t)\}) = min (\{min (\{f_1 (t), ..., f_n (t)\}), f_{n + 1} (t)\})\), because when \(f_{n + 1} (t)\) is the minimum among \(\{f_1 (t), ..., f_{n + 1} (t)\}\), \(min (\{f_1 (t), ..., f_{n + 1} (t)\}) = f_{n + 1} (t)\), while whatever the minimum among \(\{f_1 (t), ..., f_n (t)\}\) is, \(min (\{min (\{f_1 (t), ..., f_n (t)\}), f_{n + 1} (t)\}) = f_{n + 1} (t)\); when \(f_{j} (t)\) where \(j \lt n + 1\) is the minimum among \(\{f_1 (t), ..., f_{n + 1} (t)\}\), \(min (\{f_1 (t), ..., f_{n + 1} (t)\}) = f_{j} (t)\), while \(min (\{min (\{f_1 (t), ..., f_n (t)\}), f_{n + 1} (t)\}) = min (\{f_j (t), f_{n + 1} (t)\}) = f_j (t)\).
So, \(min (\{f_1, ..., f_{n + 1}\}) = min (\{min (\{f_1, ..., f_n\}), f_{n + 1}\})\).
Step 5:
Let us see that \(min (\{f_1, f_2\})\) is continuous.
Let us see that \(min (\{f_1, f_2\}) = 1 / 2 (f_1 + f_2 - \vert f_1 - f_2 \vert)\).
Let \(t \in T\) be any.
\(f_1 (t) \le f_2 (t)\) or \(f_2 (t) \lt f_1 (t)\).
Let us suppose that \(f_1 (t) \le f_2 (t)\).
\(min (\{f_1, f_2\}) (t) = min (\{f_1 (t), f_2 (t)\}) = f_1 (t)\), while \((1 / 2 (f_1 + f_2 - \vert f_1 - f_2 \vert)) (t) = 1 / 2 (f_1 (t) + f_2 (t) - \vert f_1 (t) - f_2 (t) \vert) = 1 / 2 (f_1 (t) + f_2 (t) - (f_2 (t) - f_1 (t))) = 1 / 2 (2 f_1 (t)) = f_1 (t)\).
Let us suppose that \(f_2 (t) \lt f_1 (t)\).
\(min (\{f_1, f_2\}) (t) = min (\{f_1 (t), f_2 (t)\}) = f_2 (t)\), while \((1 / 2 (f_1 + f_2 - \vert f_1 - f_2 \vert)) (t) = 1 / 2 (f_1 (t) + f_2 (t) - \vert f_1 (t) - f_2 (t) \vert) = 1 / 2 (f_1 (t) + f_2 (t) - (f_1 (t) - f_2 (t))) = 1 / 2 (2 f_2 (t)) = f_2 (t)\).
So, \(min (\{f_1, f_2\}) = 1 / 2 (f_1 + f_2 - \vert f_1 - f_2 \vert)\).
\(f_1 - f_2\) is continuous, as a well-known fact, and \(\vert f_1 - f_2 \vert\) is continuous, by the proposition that for any continuous map from any topological space into the \(1\)-dimensional Euclidean topological space, its absolute map is continuous.
Then, \(1 / 2 (f_1 + f_2 - \vert f_1 - f_2 \vert)\) is continuous, as a well-known fact.
Step 6:
Let us prove the proposition by the induction principle.
When \(n = 2\), \(min (\{f_1, ..., f_n\})\) is continuous by Step 5.
Let us suppose that for any \(n\), \(min (\{f_1, ..., f_n\})\) is continuous.
By Step 4, \(min (\{f_1, ..., f_{n + 1}\}) = min (\{min (\{f_1, ..., f_n\}), f_{n + 1}\})\), which is continuous by Step 5.
So, \(min (\{f_1, ..., f_n\})\) is continuous for each \(n \in \mathbb{N} \setminus \{0\}\).
