A description/proof of criteria for collection of open sets to be basis
Topics
About: topological space
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Description 1
- 2: Proof 1
- 3: Description 2
- 4: Proof 2
Starting Context
- The reader knows a definition of topological space.
- The reader knows a definition of basis of topological space.
- The reader admits the local criterion for openness.
Target Context
- The reader will have a description and a proof of some criteria for any collection of open sets to be a basis.
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 1
For any topological space, \(T\), any collection of open sets, \(\{B_\alpha\}\), is a basis of \(T\), if and only if each open set on \(T\) is the union of some elements of the collection.
2: Proof 1
Suppose that each open set on \(T\) is the union of some elements of the collection. For any open set, \(U\), there is the union, \(\cup_{\alpha_1} B_{\alpha_1} = U\), and any \(B_{\alpha_1}\) satisfies \(B_{\alpha_1} \subseteq U\), so, the collection is a basis.
Suppose that the collection is a basis. Any open set, \(U\), is the union of some elements of the collection, because around any point, \(p \in U\), there is an open set, \(U_p \subseteq U\), by the local criterion for openness; there is an open set, \(B_\alpha \subseteq U_p\), in the basis, by the definition of basis; \(U\) is the union of such open sets in the basis.
3: Description 2
For any set, \(S\), and any collection of subsets, \(B = \{B_\alpha \subseteq S\}\), the collection of all the unions of elements of \(B\) (the empty set is included as the union of no element) constitutes a topology for \(S\), and \(B\) is a basis of the topological space, if and only if 1) \(S\) is the union of all the elements of \(B\), which is \(S = \cup B_\alpha\), and 2) for each sets, \(B_1, B_2 \in B\), and each point, \(p \in B_1 \cap B_2\), there is a set, \(B_3 \in B\), such that \(p \in B_3 \subseteq B_1 \cap B_2\).
4: Proof 2
Suppose that the conditions 1) and 2) are satisfied. In the collection of all the unions of elements of \(B\), \(S\) is included because of 1) and the empty set is included, and any possibly uncountably infinite union of elements of the collection is included in the collection, because it is a union of elements of \(B\), and any finite intersection of elements of the collection is included in the collection, because for any elements, \(\cup_\alpha B_\alpha\) and \(\cup_\beta B_\beta\), the intersection, \((\cup_\alpha B_\alpha) \cap (\cup_\beta B_\beta) = \cup_{\alpha, \beta} (B_\alpha \cap B_\beta)\), but because of 2), for any \(p \in B_\alpha \cap B_\beta\), there is a \(B_\gamma \in B\) such that \(p \in B_\gamma \subseteq B_\alpha \cap B_\beta\), so, \(B_\alpha \cap B_\beta = \cup B_\gamma\) as such \(p\)s cover \(B_\alpha \cap B_\beta\), so, the intersection is \(\cup B_\gamma\), which is in the collection. So, the collection is indeed a topology. \(B\) is a basis of the topology, because for any open set, which is a union of elements of \(B\), and any point in the open set, the point is in an element of \(B\), and the element satisfies the definition of \(B\)'s being a basis.
Suppose that the supposed topology is really a topology and \(B\) is a basis of the topological space. By Description 1, as \(S\) is open, \(S\) is the union of some elements of \(B\), but then, \(S\) is the union of all the elements of \(B\), because adding any other remaining element, which is a subset of \(S\), does not change the union. As \(B\) is a basis, \(B_1\) and \(B_2\) are open, and \(B_1 \cap B_2\) is open, and by the definition of basis, there is such a \(B_3\).
