description/proof of that for topological space and subset, complement of interior of subset is closure of complement of subset, and complement of closure of subset is interior of complement 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 closure of subset of topological space.
- The reader admits the proposition that for any topological space and any subset, the interior of the subset is the set of the points of the space that have some open neighborhoods contained in the subset.
- 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 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.
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:
\(T \setminus Int (S) = \overline{T \setminus S}\)
\(\land\)
\(T \setminus \overline{S} = Int (T \setminus S)\)
//
2: Proof
Whole Strategy: Step 1: apply the proposition that for any topological space and any subset, the interior of the subset is the set of the points of the space that have some open neighborhoods contained in the subset to see that \(T \setminus Int (S) = \overline{T \setminus S}\); Step 2: see that \(T \setminus Int (T \setminus S) = \overline{T \setminus (T \setminus S)}\).
Step 1:
By the proposition that for any topological space and any subset, the interior of the subset is the set of the points of the space that have some open neighborhoods contained in the subset, \(Int (S) = \{t \in T \vert \exists U_t \subseteq T \in \{\text{ the open neighborhoods of } t\} (U_t \subseteq S)\}\).
So, \(T \setminus Int (S) = \{t \in T \vert \forall U_t \subseteq T \in \{\text{ the open neighborhoods of } t\} (\lnot U_t \subseteq S)\}\).
\(\lnot U_t \subseteq S\) equals \(U_t \cap (T \setminus S) \neq \emptyset\).
So, \(T \setminus Int (S) = \{t \in T \vert \forall U_t \subseteq T \in \{\text{ the open neighborhoods of } t\} (U_t \cap (T \setminus S) \neq \emptyset)\}\), which means that \(t\) is in \(T \setminus S\) or is an accumulation point of \(T \setminus S\).
So, \(T \setminus Int (S) = \overline{T \setminus S}\), 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:
\(T \setminus Int (T \setminus S) = \overline{T \setminus (T \setminus S)}\), by Step 1, \(= \overline{S}\).
That Implies that \(T \setminus \overline{S} = Int (T \setminus S)\).
