A description/proof of that finite intersection of open dense subsets of topological space is open dense
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of topological space.
- The reader knows a definition of dense subset of topological space.
Target Context
- The reader will have a description and a proof of the proposition that for any topological space, the intersection of any finite number of open dense subsets is open 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 any finite number of open dense subsets, \(\{U_i\vert i \in I\}\) where \(I\) is any finite indices set, the intersection, \(\cap_{i \in I} U_i\), is open dense.
2: Proof
\(\cap_{i \in I} U_i\) is open.
Let us suppose that \(\cap_{i \in I} U_i\) was not dense. Then, there would be a point, \(p \in T \setminus \overline{\cap_{i \in I} U_i}\). There would be an open neighborhood, \(U_p \subseteq T\), of \(p\), such that \(U_p \cap \cap_{i \in I} U_i = \emptyset\), because \(p\) would not be any accumulation point of \(\cap_{i \in I} U_i\). Let us suppose that \(I = \{1, 2, . . ., n\}\) without loss of generality. There would be a point, \(p_1 \in U_p \cap U_1\), because \(p\) would be a point of \(U_1\) or an accumulation point of \(U_1\). As \(U_1\) would be open, \(U_p \cap U_1\) would be an open neighborhood of \(p_1\), and as \(p_1\) would be a point of \(U_2\) or an accumulation point of \(U_2\), there would be a point, \(p_2 \in U_p \cap U_1 \cap U_2\), and \(U_p \cap U_1 \cap U_2\) would be an open neighborhood of \(p_2\), and so on. After all, there would be a point, \(p_n \in U_p \cap \cap_{i \in I} U_i\), a contradiction.
