A description/proof of that finite union of nowhere dense subsets of topological space has empty interior
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of nowhere dense subset of topological space.
- 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, the union of any finite number of nowhere dense subsets 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, \(T\), any finite number of nowhere dense subsets, \(\{S_i\vert i \in I\}\) where \(I\) is any finite indices set, and the union, \(S = \cup_{i \in I} S_i\), the interior of the union, \(int S\), is empty, which is \(int S = \emptyset\).
2: Proof
Let us suppose that \(int S \neq \emptyset\). There would be an open set, \(U \subseteq S\). \(\lnot U \subseteq \overline{S_1}\), because \(S_1\) would be nowhere dense. There would be a point, \(p_1 \in U\) such that \(p_1 \notin \overline{S_1}\). As \(p_1 \notin S_1\) would not be any accumulation point of \(S_1\), there would be an open neighborhood, \(U_{p_1} \subseteq T\), such that \(U_{p_1} \cap S_1 = \emptyset\). So, \(U \cap U_{p_1} \subseteq S_2 \cup S_3 \cup . . . \cup S_n\).
As \(U \cap U_{p_1}\) would play the same role for \(S_2 \cup S_3 \cup . . . \cup S_n\) as \(U\) would do for \(S_1 \cup S_2 \cup . . . \cup S_n\), there would be a neighborhood, \(U \cap U_{p_1} \cap U_{p_2} \subseteq T\) such that \(U \cap U_{p_1} \cap U_{p_2} \subseteq S_3 \cup S_4 \cup . . . \cup S_n\), and so on. After all, \(U \cap U_{p_1} \cap U_{p_2} \cap . . . \cap U_{p_{n - 1}} \subseteq S_n\), a contradiction against \(S_n\)'s being nowhere dense.
3: Note
As is shown in the proof, \(S_n\) does not really need to be nowhere dense: having the empty interior is enough, which is a weaker condition than being nowhere dense.
For an infinite union, the union may have a nonempty interior, but if \(T\) is a complete metric space or a locally compact Hausdorff space, any countable union will have the empty interior, by the Baire category theorem.
