description/proof of that range of continuous map from compact topological space into \(1\)-dimensional Euclidean topological space has minimum and maximum
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of compact topological space.
- The reader knows a definition of Euclidean topological space.
- The reader knows a definition of continuous, topological spaces map.
- The reader knows a definition of minimum of partially-ordered set.
- The reader knows a definition of maximum of partially-ordered set.
- The reader admits the proposition that for any continuous map between any topological spaces, the image of any compact subset of the domain is a compact subset of the codomain.
- The reader admits the proposition that any subset of any Euclidean metric topological space is compact if and only if the subset is closed and bounded (the Heine-Borel theorem).
- The reader admits the proposition that for the \(1\)-dimensional Euclidean topological space with the canonical linear-ordering and any subset, if the infimum of the subset exists out of the subset, the infimum is an accumulation point of the subset, and if the supremum of the subset exists out of the subset, the supremum is an accumulation point of the subset.
- The reader admits the proposition that for any topological space, any subset is closed if and only if the closure of the subset is the subset.
- The reader admits the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.
- The reader admits the proposition that for any partially-ordered set and any subset, if the infimum of the subset exists in the subset, the infimum is the minimum of the subset and if the supremum of the subset exists in the subset, the supremum is the maximum of the subset.
Target Context
- The reader will have a description and a proof of the proposition that the range of any continuous map from any compact topological space into the \(1\)-dimensional Euclidean topological space has the minimum and the maximum.
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:
\(T\): \(\in \{\text{ the compact topological spaces }\}\)
\(\mathbb{R}\): \(= \text{ the Euclidean topological space }\)
\(f\): \(: T \to \mathbb{R}\), \(\in \{\text{ the continuous maps }\}\)
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Statements:
\(\exists r, r' \in \mathbb{R} (r = Min (f (T)) \land r' = Max (f (T)))\)
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2: Proof
Whole Strategy: Step 1: see that \(f (T)\) is a compact subset of \(\mathbb{R}\) and is closed and bounded; Step 2: see that there are the infimum and the supremum of \(f (T)\); Step 3: see that the infimum and the supremum are the minimum and the maximum.
Step 1:
\(f (T)\) is compact on \(\mathbb{R}\), by the proposition that for any continuous map between any topological spaces, the image of any compact subset of the domain is a compact subset of the codomain.
\(f (T)\) is closed and bounded, by the proposition that any subset of any Euclidean metric topological space is compact if and only if the subset is closed and bounded (the Heine-Borel theorem).
Step 2:
As \(f (T)\) is bounded, there are the infimum and the supremum of \(f (T)\), \(Inf (f (T)), Sup (f (T)) \in \mathbb{R}\).
Let us suppose that \(Inf (f (T)) \notin f (T)\).
\(Inf (f (T))\) would be an accumulation point of \(f (T)\), by the proposition that for the \(1\)-dimensional Euclidean topological space with the canonical linear-ordering and any subset, if the infimum of the subset exists out of the subset, the infimum is an accumulation point of the subset, and if the supremum of the subset exists out of the subset, the supremum is an accumulation point of the subset.
But as \(f (T)\) was closed, \(f (T) = \overline{f (T)}\), by the proposition that for any topological space, any subset is closed if and only if the closure of the subset is the subset, but by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset, \(Inf (f (T)) \in \overline{f (T)} = f (T)\), because \(Inf (f (T))\) was an accumulation point, a contradiction against \(Inf (f (T)) \notin f (T)\).
So, \(Inf (f (T)) \in f (T)\).
So, \(Inf (f (T)) = Min (f (T))\), by the proposition that for any partially-ordered set and any subset, if the infimum of the subset exists in the subset, the infimum is the minimum of the subset and if the supremum of the subset exists in the subset, the supremum is the maximum of the subset.
Let us suppose that \(Sup (f (T)) \notin f (T)\).
\(Sup (f (T))\) would be an accumulation point of \(f (T)\), by the proposition that for the \(1\)-dimensional Euclidean topological space with the canonical linear-ordering and any subset, if the infimum of the subset exists out of the subset, the infimum is an accumulation point of the subset, and if the supremum of the subset exists out of the subset, the supremum is an accumulation point of the subset.
But as \(f (T)\) was closed, \(f (T) = \overline{f (T)}\), as before, but \(Sup (f (T)) \in \overline{f (T)} = f (T)\), as before, because \(Sup (f (T))\) was an accumulation point, a contradiction against \(Sup (f (T)) \notin f (T)\).
So, \(Sup (f (T)) \in f (T)\).
So, \(Sup (f (T)) = Max (f (T))\), by the proposition that for any partially-ordered set and any subset, if the infimum of the subset exists in the subset, the infimum is the minimum of the subset and if the supremum of the subset exists in the subset, the supremum is the maximum of the subset.
