A description/proof of that open sets whose complements are finite and empty set is topology
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of topological space.
- The reader admits the proposition for any set, the intersection of the compliments of any possibly uncountable number of subsets is the complement of the union of the subsets.
- The reader admits the proposition for any set, the union of the complements of any possibly uncountable number of subsets is the complement of the intersection of the subsets.
Target Context
- The reader will have a description and a proof of the proposition that for any set, the set of subsets which are subsets whose complements are finite and the empty set constitutes a topology.
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 set, \(S\), the set of subsets, \(O\), such that \(S_\alpha \in O\) if and only if \(S \setminus S_\alpha\) is finite or \(S_\alpha = \emptyset\) constitutes a topology.
2: Proof
\(S \in O\) as \(S \setminus S = \emptyset\) is finite. \(\emptyset \in O\).
Is \(\cup_\alpha S_\alpha\) open for any \(\{\alpha\}\)? \(S \setminus \cup_\alpha S_\alpha = \cap_\alpha S \setminus S_\alpha\), by the proposition for any set, the intersection of the compliments of any possibly uncountable number of subsets is the complement of the union of the subsets. As \(S \setminus S_\alpha\) is finite, \(\cap_\alpha S \setminus S_\alpha\) is finite.
Is \(\cap_i S_i\) open for any finite \(\{i\}\)? \(S \setminus \cap_i S_i = \cup_i S \setminus S_i\), by the proposition for any set, the union of the complements of any possibly uncountable number of subsets is the complement of the intersection of the subsets. As \(S \setminus S_i\) is finite and \(\{i\}\) is finite, \(\cup_i S \setminus S_i\) is finite.
