description/proof of that for continuous map from topological space into subspace of Euclidean topological space and connected subspace of domain, if component projection of image of subspace is countable, component projection is single point
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of Euclidean topological space.
- The reader knows a definition of topological subspace.
- The reader knows a definition of continuous, topological spaces map.
- The reader knows a definition of connected topological space.
- The reader admits the proposition that for any continuous map between any topological spaces, the image of any connected subspace is connected.
- The reader admits the proposition that in any nest of topological subspaces, the connected-ness of any subspace does not depend on the superspace of which the subspace is regarded to be a subspace.
- The reader admits the proposition that any subspace of any Euclidean topological space whose projection into any component is non-single countable is not connected.
Target Context
- The reader will have a description and a proof of the proposition that for any continuous map from any topological space into any subspace of any Euclidean topological space and any connected subspace of the domain, if any component projection of the image of the subspace is countable, the component projection is a single point.
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:
\(T'_1\): \(\in \{\text{ the topological spaces }\}\)
\(\mathbb{R}^d\): \(= \text{ the Euclidean topological space }\)
\(T_2\): \(\in \{\text{ the topological subspaces of } \mathbb{R}^d \}\)
\(f'\): \(: T'_1 \to T_2\), \(\in \{\text{ the continuous maps }\}\)
\(\pi_j\): \(: \mathbb{R}^d \to \mathbb{R}, (x^1, ..., x^d) \mapsto (x^j)\)
\(T_1\): \(\in \{\text{ the connected subspaces of } T'_1\}\) such that \(T_1 \neq \emptyset\)
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Statements:
\(\pi_j \circ f' (T_1) \in \{\text{ the countable sets }\}\)
\(\implies\)
\(\vert \pi_j \circ f' (T_1) \vert = 1\)
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2: Proof
Whole Strategy: Step 1: see that \(f' (T_1)\) is connected on \(T_2\) and on \(\mathbb{R}^d\); Step 2: conclude the proposition.
Step 1:
\(f' (T_1)\) is connected as the subspace of \(T_2\), by the proposition that for any continuous map between any topological spaces, the image of any connected subspace is connected.
\(f' (T_1)\) is connected as the subspace of \(\mathbb{R}^d\), by the proposition that in any nest of topological subspaces, the connected-ness of any subspace does not depend on the superspace of which the subspace is regarded to be a subspace.
Step 2:
\(\pi_j (f' (T_1))\) is countable by the supposition.
\(0 \lt \vert \pi_j (f' (T_1)) \vert\), because \(T_1 \neq \emptyset\).
Let us suppose that \(1 \lt \vert \pi_j (f' (T_1)) \vert\).
\(f' (T_1)\) would not be connected, by the proposition that any subspace of any Euclidean topological space whose projection into any component is non-single countable is not connected.
So, \(\vert \pi_j (f' (T_1)) \vert \le 1\).
So, \(\vert \pi_j (f' (T_1)) \vert = 1\).
