description/proof of that for topological space and subspace with some connected-components removed, remained connected-components are connected-components of subspace
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of connected topological component.
- The reader knows a definition of topological subspace.
- 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 connected topological component is exactly any connected topological subspace that cannot be made larger.
Target Context
- The reader will have a description and a proof of the proposition that for any topological space and the subspace with any some connected-components removed, the remained connected-components are the connected-components of the subspace.
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:
\(T'\): \(\in \{\text{ the topological spaces }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(S'\): \(= \{C_j \in \{\text{ the connected components of } T'\} \vert j \in J\}\)
\(S\): \(\subset S'\)
\(T\): \(= \cup S \subset T'\), with the subspace topology
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Statements:
\(S = \{\text{ the connected components of } T\}\)
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2: Proof
Whole Strategy: Step 1: see that each \(C_j \in S\) is connected on \(T\); Step 2: see that for each \(j \neq l\), each point in \(C_j\) is not connected to any point in \(C_l\) on \(T\); Step 3: conclude the proposition.
Step 1:
Each \(C_j \in S\) is connected on \(T\), 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:
Let us see that for each \(j, l \in J\) such that \(j \neq l\), each point in \(C_j\) is not connected to any point in \(C_l\) on \(T\).
Let us suppose that a \(c_j \in C_j\) was connected to a \(c_l \in C_l\) on \(T\).
There would be a connected subspace of \(T\), \(T^` \subseteq T\), such that \(c_j, c_l \in T^`\).
\(T^`\) would be a connected subspace of \(T'\), 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, which would mean that \(c_j\) and \(c_l\) were connected on \(T'\), a contradiction against that \(c_j\) and \(c_l\) were not in the same connected-component of \(T'\).
So, for each \(j, l \in J\) such that \(j \neq l\), each point in \(C_j\) is not connected to any point in \(C_l\) on \(T\).
Step 3:
So, each \(C_j\) is a connected subspace of \(T\) that cannot be made larger, so, \(C_j\) is a connected component of \(T\), by the proposition that any connected topological component is exactly any connected topological subspace that cannot be made larger.
