A description/proof of that for subset of topological space, closure of subset minus subset has empty interior
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of closure of subset.
- The reader knows a definition of interior of subset.
Target Context
- The reader will have a description and a proof of the proposition that for any topological space and any its subset, the closure of the subset minus the subset has the empty interior.
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 any subset, \(S \subseteq T\), the interior of \(\overline{S} \setminus S\) is empty, which is \(int (\overline{S} \setminus S) = \emptyset\).
2: Proof
For any point, \(p \in \overline{S} \setminus S\), \(p\) is an accumulation point of \(S\), and so, any neighborhood of \(p\), \(N_p \subseteq T\), intersects \(S\), which means that \(N_p\) is not contained in \(\overline{S} \setminus S\). So, there is no open set contained in \(\overline{S} \setminus S\).
3: Note
When \(S\) is open on \(T\), \(\overline{S} \setminus S\) is closed on \(T\), by the proposition that any closed set minus any open set is closed, so, \(\overline{\overline{S} \setminus S} = \overline{S} \setminus S\), so, \(int (\overline{\overline{S} \setminus S}) = \emptyset\), which means that \(\overline{S} \setminus S\) is nowhere dense.
