A description/proof of that complement of empty-interior especially nowhere dense subset is dense
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of nowhere dense subset.
- The reader knows a definition of dense subset.
- The reader admits the proposition that for any subset of any topological space, the closure of the complement of the subset is the complement of the interior of the subset.
Target Context
- The reader will have a description and a proof of the proposition that for any topological space and its any empty-interior especially nowhere dense subset, the complement of the subset is dense.
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: Description
For any topological space, \(T\), and its any empty-interior especially nowhere dense subset, \(S \subseteq T\), which means that \(int S = \emptyset\) especially \(int \overline{S} = \emptyset\), the complement of \(S\), \(T \setminus S\), is dense, which means that \(\overline{T \setminus S} = T\).
2: Proof
\(\overline{T \setminus S} = T \setminus int S\), by the proposition that for any subset of any topological space, the closure of the complement of the subset is the complement of the interior of the subset. As \(int S = \emptyset\), \(\overline{T \setminus S} = T\). When \(int \overline{S} = \emptyset\), \(int S = \emptyset\), because as \(\overline{S}\) does not contain any open subset of \(T\), the possibly smaller \(S\) does not contain any open subset of \(T\).
