description/proof of that for topological space and \(2\) connected subspaces, connected-components of union of subspaces are subspaces or union of subspaces
Topics
About: topological space
The table of contents of this article
Starting Context
Target Context
- The reader will have a description and a proof of the proposition that for any topological space and any \(2\) connected subspaces, the connected-components of the union of the subspaces are the subspaces or the union of the subspaces.
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:
\(T'\): \(\in \{\text{ the topological spaces }\}\)
\(T_1\): \(\in \{\text{ the connected subspaces of } T'\}\)
\(T_2\): \(\in \{\text{ the connected subspaces of } T'\}\)
\(T_1 \cup T_2\): \(\subseteq T'\) as the topological subspace
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Statements:
\(T_1 \cup T_2\) has the connected-components, \(\{T_1, T_2\}\) or \(\{T_1 \cup T_2\}\)
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2: Note
Refer to the proposition that for any topological space, the union of any 2 connected subspaces is connected if each neighborhood of a point on one of the subspaces contains a point of the other subspace and the proposition that for any topological space, the union of any connected subspaces that share any point is connected.
3: Proof
Whole Strategy: Step 1: see that for each \(t_1, t'_1 \in T_1\), \(t_1\) and \(t'_1\) are connected on \(T_1 \cup T_2\) and for each \(t_2, t'_2 \in T_2\), \(t_2\) and \(t'_2\) are connected on \(T_1 \cup T_2\); Step 2: suppose that each \(t_1\) is not connected to any \(t_2\), and see that \(\{T_1, T_2\}\) are the connected components; Step 3: suppose that a \(t_1\) is connected to a \(t_2\), and see that \(\{T_1 \cup T_2\}\) is the connected component.
Step 1:
Let \(t_1, t'_1 \in T_1\) be any.
\(T_1\) as the subspace of \(T'\) is the subspace of \(T_1 \cup T_2\), by the proposition that in any nest of topological subspaces, the openness of any subset on any subspace does not depend on the superspace of which the subspace is regarded to be a subspace.
So, \(T_1\) is a connected subspace of \(T_1 \cup T_2\) that contains \(t_1\) and \(t'_1\).
So, \(t_1\) and \(t'_1\) are connected on \(T_1 \cup T_2\).
For each \(t_2, t'_2 \in T_2\), \(t_2\) and \(t'_2\) are connected on \(T_1 \cup T_2\).
Step 2:
There are only the 2 cases: 1) each \(t_1 \in T_1\) is not connected to any \(t_2 \in T_2\); 2) a \(t_1 \in T_1\) is connected to a \(t_2 \in T_2\).
Let us suppose that 1) each \(t_1 \in T_1\) is not connected to any \(t_2 \in T_2\).
\(T_1\) is an equivalence class of connected-ness on \(T_1 \cup T_2\), because each pair of points on \(T_1\) are connected by Step 1 and each point not on \(T_1\) is on \(T_2\) and does not belong to the equivalence class, by the supposition.
\(T_2\) is an equivalence class of connected-ness on \(T_1 \cup T_2\), likewise.
So, \(\{T_1, T_2\}\) are the connected components of \(T_1 \cup T_2\).
Step 3:
Let us suppose that 2) a \(t_1 \in T_1\) is connected to a \(t_2 \in T_2\).
Each pair of points on \(T_1 \cup T_2\) are connected, because when they are \(t'_1, t''_1 \in T_1\), they are connected by Step 1; when they are \(t'_1 \in T_1\) and \(t'_2 \in T_2\), they are connected because \(t'_1\) and \(t_1\) are connected, \(t_1\) and \(t_2\) are connected, and \(t_2\) and \(t'_2\) are connected; when they are \(t'_2, t''_2 \in T_2\), they are connected by Step 1.
So, \(T_1 \cup T_2\) is the equivalence class of connected-ness on \(T_1 \cup T_2\).
So, \(\{T_1 \cup T_2\}\) is the connected component of \(T_1 \cup T_2\).
