description/proof of that for topological space, union of connected subspaces that share point is connected
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, the union of any connected subspaces that share any point is connected.
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:
\(T'\): \(\in \{\text{ the topological spaces }\}\)
\(t'\): \(\in T'\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{T_j \in \{\text{ the connected subspaces of } T'\}\vert j \in J\}\): such that \(t' \in T_j\)
\(T\): \(= \cup_{j \in J} T_j\)
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Statements:
\(T \in \{\text{ the connected subspaces of } T'\}\)
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2: Proof
Whole Strategy: see it by a reduction to absurdity; Step 1: suppose that \(T\) was not connected as \(T = U_1 \cup U_2\); Step 2: suppose that \(t' \in U_1\) and see that \(T_j \subseteq U_1\); Step 3: conclude the proposition.
Step 1:
Let us suppose that \(T\) was not connected.
There would be some nonempty open \(U_1, U_2 \subseteq T\) such that \(T = U_1 \cup U_2\) and \(U_1 \cap U_2 = \emptyset\).
Step 2:
Let us suppose that \(t' \in U_1\) without loss of generality.
For each \(j \in J\), let us see that \(T_j \subseteq U_1\).
\(T_j = T_j \cap (U_1 \cap U_2)\).
\(= (T_j \cap U_1) \cup (T_j \cap U_2)\), by the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset.
\(T_j \cap U_1 \neq \emptyset\), because \(t'\) was there.
\(T_j \cap U_1\) would be open on \(T_j\), because as \(U_1\) was open on \(T\), \(U_1 = U'_1 \cap T\) for an open subset of \(T'\), \(U'_1 \subseteq T'\), but \(T_j \cap U_1 = T_j \cap U'_1 \cap T = (T_j \cap T) \cap U'_1 = T_j \cap U'_1\).
Likewise, \(T_j \cap U_2\) would be open on \(T_j\).
\((T_j \cap U_1) \cap (T_j \cap U_2) = \emptyset\), because \(U_1 \cap U_2 = \emptyset\).
So, as \(T_j\) was connected, \(T_j \cap U_2 = \emptyset\).
That would mean that \(T_j \subseteq U_1\), because while \(T_j \subseteq T = U_1 \cup U_2\), each point on \(T_j\) would not be on \(U_2\), so, would be on \(U_1\).
Step 3:
So, \(T = \cup_{j \in J} T_j \subseteq U_1\).
So, \(U_2 = \emptyset\), because otherwise, any point on \(U_2\) would not be on \(T\), because the point would not be on \(U_1\).
That is a contradiction against the supposition, so, the supposition was wrong, and \(T\) is connected.
