description/proof of that for set augmented with set as an element, set of subsets of set and subsets whose complements are finite 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, for the set augmented with the set as an element, the set of the subsets of the set and the subsets whose complements are finite 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: Structured Description
Here is the rules of Structured Description.
Entities:
\(S\): \(\in \{\text{ the sets }\}\)
\(S'\): \(= S \cup \{S\}\)
\(O\): \(\subseteq Pow (S')\), \(= \{\text{ the subsets of } S\} \cup \{S^` \subseteq S' \vert S' \setminus S^` \in \{\text{ the finite sets }\}\}\)
//
Statements:
\(O \in \{\text{ the topologies for } S'\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(S', \emptyset \in O\); Step 2: see that the union of any subset of \(O\) is in \(O\); Step 3: see that the intersection of any finite subset of \(O\) is in \(O\); Step 4: conclude the proposition.
Step 1:
\(S' \in O\), because \(S' \setminus S' = \emptyset\) is finite.
\(\emptyset \in O\), because \(\emptyset \subseteq S\).
Step 2:
Let \(\{U_j \vert j \in J\} \subseteq O\) be any where \(J\) is a possibly uncountable index set.
For each \(j \in J\), \(U_j \in \{\text{ the subsets of } S\}\) or \(U_j \in \{S^` \subseteq S' \vert S' \setminus S^` \in \{\text{ the finite sets }\}\}\).
When for each \(j \in J\), \(U_j \in \{\text{ the subsets of } S\}\), \(\cup_{j \in J} U_j \in \{\text{ the subsets of } S\}\), so, \(\cup_{j \in J} U_j \in O\).
Let us suppose that there is a \(j' \in J\) such that \(U_{j'} \notin \{\text{ the subsets of } S\}\).
\(U_{j'} \in \{S^` \subseteq S' \vert S' \setminus S^` \in \{\text{ the finite sets }\}\}\), so, \(S' \setminus U_{j'}\) is finite.
\(S' \setminus \cup_{j \in J} U_j = \cap_{j \in J} (S' \setminus U_j)\), 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, \(\subseteq S' \setminus U_{j'}\), which is finite.
So, \(\cup_{j \in J} U_j \in \{S^` \subseteq S' \vert S' \setminus S^` \in \{\text{ the finite sets }\}\}\), so, \(\cup_{j \in J} U_j \in O\).
So, anyway, \(\cup_{j \in J} U_j \in O\).
Step 3:
Let \(\{U_j \vert j \in J\} \subseteq O\) be any where \(J\) is a finite index set.
For each \(j \in J\), \(U_j \in \{\text{ the subsets of } S\}\) or \(U_j \in \{S^` \subseteq S' \vert S' \setminus S^` \in \{\text{ the finite sets }\}\}\).
When there is a \(j' \in J\) such that \(U_{j'} \in \{\text{ the subsets of } S\}\), \(\cap_{j \in J} U_j \subseteq U_{j'} \in \{\text{ the subsets of } S\}\), so, \(\cap_{j \in J} U_j \in O\).
Let us suppose that for each \(j \in J\), \(U_j \notin \{\text{ the subsets of } S\}\).
For each \(j \in J\), \(U_j \in \{S^` \subseteq S' \vert S' \setminus S^` \in \{\text{ the finite sets }\}\}\).
\(S' \setminus \cap_{j \in J} U_j = \cup_{j \in J} (S' \setminus U_j)\), 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, which is finite, because each \(S' \setminus U_j\) is finite and \(J\) is finite.
So, \(\cap_{j \in J} U_j \in \{S^` \subseteq S' \vert S' \setminus S^` \in \{\text{ the finite sets }\}\}\), so, \(\cap_{j \in J} U_j \in O\).
So, anyway, \(\cap_{j \in J} U_j \in O\).
Step 4:
So, \(O\) is a topology for \(S'\).
