description/proof of that for topological space, boundary of intersection of locally finite set of subsets is contained in union of boundaries of subsets
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of boundary of subset of topological space.
- The reader admits the proposition that for any set, the union of the complements of any possibly uncountable number of subsets is the complement of the intersection of the subsets.
- The reader admits the proposition that for any locally finite set of subsets of any topological space, the closure of the union of the subsets is the union of the closures of the subsets.
- The reader admits the proposition that for any topological space, the closure of the intersection of any subsets is contained in the intersection of the closures of the subsets.
Target Context
- The reader will have a description and a proof of the proposition that for any topological space, the boundary of the intersection of any possibly uncountable locally finite set of subsets is contained in the union of the boundaries 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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{S_j \subseteq T \vert j \in J\}\): \(\in \{\text{ the locally finite sets of subsets of } T\}\)
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Statements:
\(Bou (\cap_{j \in J} S_j) \subseteq \cup_{j \in J} Bou (S_j)\)
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2: Note
Equality does not necessarily hold even if \(J\) is finite.
For example, let \(T = \mathbb{R}\) with the Euclidean topology and \(\{S_1 = (-1, 0), S_2 = (0, 1)\}\), then, \(Bou (\cap_{j \in J} S_j) = Bou ((-1, 0) \cap (0, 1)) = Bou (\emptyset) = \emptyset \subset \cup_{j \in J} Bou (S_j) = Bou ((-1, 0)) \cup Bou ((0, 1)) = \{-1, 0\} \cap \{0, 1\} = \{-1, 0, 1\}\).
3: Proof
Whole Strategy: Step 1: note that when \(J\) is infinite, \(Bou (\cap_{j \in J} S_j) = \emptyset\); Step 2: when \(J\) is finite, let \(s \in Bou (\cap_{j \in J} S_j)\) be any and see that \(s \in \cup_{j \in J} Bou (S_j)\).
Step 1:
While this proposition says that \(J\) is possibly infinite and \(\{S_j \subseteq T \vert j \in J\}\) is locally finite, in fact, when \(J\) is infinite, \(\cap_{j \in J} S_j = \emptyset\), because if \(s \in \cap_{j \in J} S_j\), any neighborhood of \(s\) would intersect each \(S_j\), so, \(\{S_j \subseteq T \vert j \in J\}\) would not be locally finite, a contradiction, so, there is no such any \(s\).
Then, \(Bou (\cap_{j \in J} S_j) = Bou (\emptyset) = \emptyset\).
Then, \(Bou (\cap_{j \in J} S_j) \subseteq \cup_{j \in J} Bou (S_j)\).
Step 2:
Let us suppose that \(J\) is finite.
Let \(s \in Bou (\cap_{j \in J} S_j)\) be any.
\(Bou (\cap_{j \in J} S_j) = \overline{\cap_{j \in J} S_j} \cap \overline{T \setminus \cap_{j \in J} S_j}\), by definition.
\(T \setminus \cap_{j \in J} S_j = \cup_{j \in J} (T \setminus S_j)\), by the proposition for any set, the union of the complements of any possibly uncountable number of subsets is the complement of the intersection of the subsets.
\(\overline{\cup_{j \in J} (T \setminus S_j)} = \cup_{j \in J} \overline{T \setminus S_j}\), by the proposition that for any locally finite set of subsets of any topological space, the closure of the union of the subsets is the union of the closures of the subsets: \(\{T \setminus S_j \vert j \in J\}\) is finite, so, locally finite.
So, \(Bou (\cap_{j \in J} S_j) = \overline{\cap_{j \in J} S_j} \cap \cup_{j \in J} \overline{T \setminus S_j}\).
On the other hand, \(\cup_{j \in J} Bou (S_j) = \cup_{j \in J} (\overline{S_j} \cap \overline{T \setminus S_j})\).
As \(\overline{\cap_{j \in J} S_j} \subseteq \cap_{j \in J} \overline{S_j}\) by the proposition that for any topological space, the closure of the intersection of any subsets is contained in the intersection of the closures of the subsets, while \(s \in \overline{\cap_{j \in J} S_j}\), for each \(j \in J\), \(s \in \overline{S_j}\), and as \(s \in \cup_{j \in J} \overline{T \setminus S_j}\), \(s \in \overline{T \setminus S_j}\) for a \(j \in J\), so, \(s \in \overline{S_j} \cap \overline{T \setminus S_j}\) for a \(j \in J\), so, \(s \in \cup_{j \in J} (\overline{S_j} \cap \overline{T \setminus S_j}) = \cup_{j \in J} Bou (S_j)\).
So, \(s \in \cup_{j \in J} Bou (S_j)\).
