description/proof of that for continuous map from connected topological space into \(1\)-dimensional Euclidean topological space, if range contains \(2\) points, range contains any point between points
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of connected topological space.
- 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 map, the map preimage of any union of sets is the union of the map preimages of the sets.
- The reader admits the proposition that the preimages of any disjoint subsets under any map are disjoint.
Target Context
- The reader will have a description and a proof of the proposition that for any continuous map from any connected topological space into the \(1\)-dimensional Euclidean topological space, if the range contains any \(2\) points, the range contains any point between the points.
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:
\(T_1\): \(\in \{\text{ the connected topological spaces }\}\)
\(\mathbb{R}\): \(= \text{ the Euclidean topological space }\)
\(f\): \(: T_1 \to \mathbb{R}\), \(\in \{\text{ the continuous maps }\}\)
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Statements:
\(\exists r_1, r_2 \in f (T_1)\)
\(\implies\)
\(\forall t \in [0, 1] (r_1 + t (r_2 - r_1) \in f (T_1))\)
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2: Proof
Whole Strategy: Step 1: suppose that \(r_1 + t (r_2 - r_1) \notin f (T_1)\) and find a contradiction.
Step 1:
Let us suppose that \(r_1 + t (r_2 - r_1) \notin f (T_1)\).
\((- \infty, r_1 + t (r_2 - r_1)) \subseteq \mathbb{R}\) and \((r_1 + t (r_2 - r_1), \infty) \subseteq \mathbb{R}\) would be some open subsets of \(\mathbb{R}\).
\(f^{-1} (\mathbb{R} \setminus \{r_1 + t (r_2 - r_1)\}) = f^{-1} ((- \infty, r_1 + t (r_2 - r_1)) \cup (r_1 + t (r_2 - r_1), \infty)) = T_1\).
But \(f^{-1} ((- \infty, r_1 + t (r_2 - r_1)) \cup (r_1 + t (r_2 - r_1), \infty)) = f^{-1} ((- \infty, r_1 + t (r_2 - r_1))) \cup f^{-1} ((r_1 + t (r_2 - r_1), \infty))\), by the proposition that for any map, the map preimage of any union of sets is the union of the map preimages of the sets.
\(f^{-1} ((- \infty, r_1 + t (r_2 - r_1))) \subseteq T_1\) and \(f^{-1} ((r_1 + t (r_2 - r_1), \infty)) \subseteq T_1\) would be some open subsets of \(T_1\), because \(f\) was continuous.
\(f^{-1} ((- \infty, r_1 + t (r_2 - r_1))) \neq \emptyset\) and \(f^{-1} ((r_1 + t (r_2 - r_1), \infty)) \neq \emptyset\), because \(r_1, r_2 \in f (T_1)\).
\(f^{-1} ((- \infty, r_1 + t (r_2 - r_1))) \cap f^{-1} ((r_1 + t (r_2 - r_1), \infty)) = \emptyset\), by the proposition that the preimages of any disjoint subsets under any map are disjoint.
So, \(T_1\) would be disconnected, a contradiction.
So, \(r_1 + t (r_2 - r_1) \notin f (T_1)\) was wrong, and so, \(r_1 + t (r_2 - r_1) \in f (T_1)\).
