description/proof of that for topological space and open cover, subset is open iff intersection of subset and each element of open cover is open
Topics
About: topological space
The table of contents of this article
Starting Context
Target Context
- The reader will have a description and a proof of the proposition that for any topological space and any open cover, any subset is open iff the intersection of the subset and each element of the open cover is open.
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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(A\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{U_\alpha \vert \alpha \in A\}\): \(\in \{\text{ the open covers of } T\}\)
\(S \subseteq T\):
//
Statements:
\(S \in \{\text{ the open subsets of } T\}\)
\(\iff\)
\(\forall \alpha \in A (S \cap U_\alpha \in \{\text{ the open subsets of } T\})\)
//
2: Note
This proposition says "Openness of subset can be checked locally." so to speak.
3: Proof
Whole Strategy: Step 1: suppose that \(S\) is open and see that \(S \cap U_\alpha\) is open; Step 2: suppose that \(S \cap U_\alpha\) is open, see that \(S = T \cap S = (\cup_{\alpha \in A} U_\alpha) \cap S\), and see that \(S\) is open.
Step 1:
Let us suppose that \(S\) is open on \(T\).
For each \(\alpha \in A\), \(S \cap U_\alpha\) is open on \(T\) as a finite intersection of open subsets.
Step 2:
Let us suppose that for each \(\alpha \in A\), \(S \cap U_\alpha\) is open on \(T\).
\(S = T \cap S = (\cup_{\alpha \in A} U_\alpha) \cap S = \cup_{\alpha \in A} (U_\alpha \cap S)\), by the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset, which is open on \(T\) as a union of open subsets.
