description/proof of that for topological space, union of closures of subsets is contained in closure of union of subsets
Topics
About: topological space
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
Starting Context
Target Context
- The reader will have a description and a proof of the proposition that for any topological space and any set of any subsets, the union of the closures of the subsets is contained in 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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(\{S_\beta \subseteq T \vert \beta \in B\}\): \(B \in \{\text{ the possibly uncountable index sets }\}\)
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Statements:
\(\cup_{\beta \in B} \overline{S_\beta} \subseteq \overline{\cup_{\beta \in B} S_\beta}\), where the overlines denote the closures.
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2: Natural Language Description
For any topological space, \(T\), and any set of any subsets, \(\{S_\beta \subseteq T \vert \beta \in B\}\), where \(B\) is any possibly uncountable index set, \(\cup_{\beta \in B} \overline{S_\beta} \subseteq \overline{\cup_{\beta \in B} S_\beta}\), where the overlines denote the closures.
3: Proof
Whole Strategy: Step 1: for each \(p \in \cup_{\beta \in B} \overline{S_\beta}\), take a \(\beta \in B\) such that \(p \in \overline{S_\beta}\), and list the 2 cases, Case 1: \(p \in S_\beta\); Case 2: \(p \notin S_\beta\) and \(p\) is an accumulation point of \(S_\beta\); Step 2: for Case 1, see that \(p \in \overline{\cup_{\beta \in B} S_\beta}\); Step 3: for Case 2, see that \(p \in \overline{\cup_{\beta \in B} S_\beta}\); Step 4: conclude the proposition.
Step 1:
For each \(p \in \cup_{\beta \in B} \overline{S_\beta}\), \(p \in \overline{S_\beta}\) for a \(\beta \in B\).
\(p \in S_\beta\) or (\(p \notin S_\beta\) and \(p\) is an accumulation point of \(S_\beta\)), by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.
Step 2:
When \(p \in S_\beta\), \(p \in \cup_{\beta \in B} S_\beta\), so, \(p \in \overline{\cup_{\beta \in B} S_\beta}\).
Step 3:
When \(p \notin S_\beta\) and \(p\) is an accumulation point of \(S_\beta\), for each neighborhood of \(p\), \(N_p \subseteq T\), \(N_p \cap S_\beta \neq \emptyset\), so, \(N_p \cap (\cup_{\beta \in B} S_\beta) \neq \emptyset\), so, \(p\) is an accumulation point of \(\cup_{\beta \in B} S_\beta\). So, \(p \in \overline{\cup_{\beta \in B} S_\beta}\).
Step 4:
So, \(\cup_{\beta \in B} \overline{S_\beta} \subseteq \overline{\cup_{\beta \in B} S_\beta}\).
