A description/proof of that complement of open dense subset is nowhere 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 interior of the complement of the subset is the complement of the closure 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 open dense subset, the complement of the subset is nowhere 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 open dense subset, \(S \subseteq T\), which means that \(\overline{S} = T\), the complement of \(S\), \(T \setminus S\), is nowhere dense, which means that \(int (\overline{T \setminus S}) = \emptyset\).
2: Proof
\(int (T \setminus S) = T \setminus \overline{S}\), by the proposition that for any subset of any topological space, the interior of the complement of the subset is the complement of the closure of the subset. As \(\overline{S} = T\), \(int (T \setminus S) = \emptyset\). As \(S\) is open, \(T \setminus S\) is closed, and \(\overline{T \setminus S} = T \setminus S\), so, \(int (\overline{T \setminus S}) = int (T \setminus S) = \emptyset\).
