description/proof of that for disjoint union topological space, closure of disjoint union of subsets is disjoint union of closures of subsets
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of disjoint union topology.
- The reader knows a definition of closure of subset of topological space.
- The reader admits the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.
Target Context
- The reader will have a description and a proof of the proposition that for any disjoint union topological space, the closure of the disjoint union of any subsets is the disjoint union of the closures of the subsets.
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:
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{T_j \vert j \in J\}\): \(T_j \in \{\text{ the topological spaces }\}\)
\(T\): \(= \coprod_{j \in J} T_j\)
\(\{S_j \subseteq T_j \vert j \in J\}\):
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Statements:
\(\overline{\coprod_{j \in J} S_j} = \coprod_{j \in J} \overline{S_j}\), where the 1st overline is the closure on \(T\) and the 2nd overline is the closure on \(T_j\)
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2: Proof
Whole Strategy: Step 1: see that \(\overline{\coprod_{j \in J} S_j} \subseteq \coprod_{j \in J} \overline{S_j}\); Step 2: see that \(\coprod_{j \in J} \overline{S_j} \subseteq \overline{\coprod_{j \in J} S_j}\).
Step 1:
\(\coprod_{j \in J} S_j \subseteq \coprod_{j \in J} \overline{S_j}\), and the right hand side is closed on \(T\), by the definition of disjoint union topology.
So, \(\overline{\coprod_{j \in J} S_j} \subseteq \coprod_{j \in J} \overline{S_j}\), by the definition of closure of subset of topological space.
Step 2:
Let \(p \in \coprod_{j \in J} \overline{S_j}\) be any.
\(p \in \overline{S_j}\) for a \(j\).
Let \(U_p \subseteq T\) be any open neighborhood of \(p\).
\(U_p \cap T_j\) is open on \(T_j\), by the definition of disjoint union topology, and \(p \in U_p \cap T_j\).
So, \(U_p \cap T_j\) is an open neighborhood of \(p\) on \(T_j\).
\(U_p \cap T_j \cap S_j \neq \emptyset\), because \(p \in \overline{S_j}\), by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.
So, \(U_p \cap \coprod_{j \in J} S_j \neq \emptyset\).
So, \(p \in \overline{\coprod_{j \in J} S_j}\), by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.
So, \(\coprod_{j \in J} \overline{S_j} \subseteq \overline{\coprod_{j \in J} S_j}\).
So, \(\overline{\coprod_{\alpha \in A} S_\alpha} = \coprod_{\alpha \in A} \overline{S_\alpha}\).
