description/proof of that for topological space and subsets, closure of union of closures of subsets is closure of union of subsets
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 subsets, the closure of the union of the closures of the subsets is the closure of the union 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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{S_j \subseteq T \vert j \in J\}\):
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Statements:
\(\overline{\cup_{j \in J} \overline{S_j}} = \overline{\cup_{j \in J} S_j}\)
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2: Proof
Whole Strategy: apply the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset; Step 1: see that \(\overline{\cup_{j \in J} \overline{S_j}} \subseteq \overline{\cup_{j \in J} S_j}\); Step 2: see that \(\overline{\cup_{j \in J} S_j} \subseteq \overline{\cup_{j \in J} \overline{S_j}}\); Step 3: conclude the proposition.
Step 1:
Let \(p \in \overline{\cup_{j \in J} \overline{S_j}}\) be any.
Let \(U_p \subseteq T\) be any open neighborhood of \(p\).
\(U_p \cap \cup_{j \in J} \overline{S_j} \neq \emptyset\), by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.
So, for a \(j \in J\), \(U_p \cap \overline{S_j} \neq \emptyset\).
But that means that \(U_p \cap S_j \neq \emptyset\), because otherwise, \(S_j \subseteq \overline{S_j} \cap (T \setminus U_p) \subset \overline{S_j}\), because \(S_j \subseteq T \setminus U_p\), so, \(S_j \subseteq \overline{S_j} \cap (T \setminus U_p) \subseteq \overline{S_j}\), but as \(U_p \cap \overline{S_j} \neq \emptyset\), \(T \setminus U_p\) lacked a point of \(\overline{S_j}\), so, \(\overline{S_j} \cap (T \setminus U_p) \subset \overline{S_j}\), a contradiction against that \(\overline{S_j}\) was the intersection of the closed subsets that contained \(S_j\): \(\overline{S_j} \cap (T \setminus U_p)\) was closed.
So, \(U_p \cap \cup_{j \in J} S_j \neq \emptyset\), so, \(p \in \overline{\cup_{j \in J} S_j}\), by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.
So, \(\overline{\cup_{j \in J} \overline{S_j}} \subseteq \overline{\cup_{j \in J} S_j}\).
Step 2:
\(\overline{\cup_{j \in J} S_j} \subseteq \overline{\cup_{j \in J} \overline{S_j}}\), because \(\cup_{j \in J} S_j \subseteq \cup_{j \in J} \overline{S_j}\): each closed subset that contains \(\cup_{j \in J} \overline{S_j}\) contains \(\cup_{j \in J} S_j\).
Step 3:
So, \(\overline{\cup_{j \in J} \overline{S_j}} = \overline{\cup_{j \in J} S_j}\).
