description/proof of that for topological space and subset, space is disjoint union of interior of subset, boundary of subset, and complement of closure of subset
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of interior of subset of topological space.
- The reader knows a definition of boundary of subset of topological space.
- The reader knows a definition of closure of subset of topological space.
- The reader admits the proposition that for any topological space and any subset, the complement of the interior of the subset is the closure of the complement of the subset, and the complement of the closure of the subset is the interior of the complement of the subset.
Target Context
- The reader will have a description and a proof of the proposition that for any topological space and any subset, the space is the disjoint union of the interior of the subset, the boundary of the subset, and the complement of the closure of the subset.
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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(S\): \(\subseteq T\)
//
Statements:
\(\{Int (S), Bou (S), T \setminus \overline{S}\} \in \{\text{ the disjoint sets of subsets }\}\)
\(\land\)
\(T = Int (S) \cup Bou (S) \cup (T \setminus \overline{S})\)
//
2: Proof
Whole Strategy: Step 1: see that \(\{Int (S), Bou (S), T \setminus \overline{S}\}\) is disjoint; Step 2: see that for each \(t \in T\), \(t \in Int (S) \cup Bou (S) \cup (T \setminus \overline{S})\).
Step 1:
Let us see that \(\{Int (S), Bou (S), T \setminus \overline{S}\}\) is disjoint.
Let \(t \in Int (S)\) be any.
\(Int (S) = T \setminus \overline{T \setminus S}\), by the proposition that for any topological space and any subset, the complement of the interior of the subset is the closure of the complement of the subset, and the complement of the closure of the subset is the interior of the complement of the subset.
So, \(t \notin \overline{T \setminus S}\), so, \(t \notin \overline{S} \cap \overline{T \setminus S} = Bou (S)\).
As \(Int (S) \subseteq S \subseteq \overline{S}\), \(t \notin T \setminus \overline{S}\).
So, \(Int (S)\) is disjoint with \(Bou (S)\) and with \(T \setminus \overline{S}\).
We already know that \(Bou (S)\) is disjoint with \(Int (S)\).
\(Bou (S) = \overline{S} \cap \overline{T \setminus S} \subseteq \overline{S}\).
So, \(Bou (S)\) is disjoint with \(T \setminus \overline{S}\).
We already know that \(T \setminus \overline{S}\) is disjoint with \(Int (S)\) and with \(Bou (S)\).
So, \(\{Int (S), Bou (S), T \setminus \overline{S}\}\) is disjoint.
Step 2:
Let \(t \in T\) be any.
Let us suppose that \(t \notin \overline{S}\).
Then, \(t \in T \setminus \overline{S}\).
Let us suppose that \(t \in \overline{S}\).
When \(t \notin Int (S)\), as \(Int (S) = T \setminus \overline{T \setminus S}\), \(t \in \overline{T \setminus S}\), so, \(t \in \overline{S} \cap \overline{T \setminus S} = Bou (S)\).
So, \(t \in Int (S) \cup Bou (S)\).
So, in any case, \(t \in Int (S) \cup Bou (S) \cup (T \setminus \overline{S})\).
