Munkres - Basis for a Topology

Adar Kahiri
July 2020

Problem 4:


(a) All we must do is show that the intersection of the topologies has the properties of a topology:
  1. \(\varnothing\) and \(X\) are clearly in \(\bigcap \mathscr{T}_{\alpha}\) since they are both in every \(\mathscr{T}_{\alpha}\) by definition.
  2. Let \(x, y \in \bigcap \mathscr{T}_{\alpha}\). Since both \(x\) and \(y\) are contained in every \(\mathscr{T}_{\alpha}\), their union is also contained in every \(\mathscr{T}_{\alpha}\) by definition.
  3. Since both \(x\) and \(y\) are contained in every \(\mathscr{T}_{\alpha}\), their intersection is also contained in every \(\mathscr{T}_{\alpha}\) by definition.
The same is not true for \(\bigcup \mathscr{T}_{\alpha}\) unless there is one \(\mathscr{T}_{\alpha}\) that contains all the other topologies. For example, suppose you have \(x \in \mathscr{T}_{\alpha}\) and \(y \in \mathscr{T}_{\beta}\). Unless \(y \in \mathscr{T}_{\alpha}\) or \(x \in \mathscr{T}_{\beta}\), \(x \cup y\) isn't necessarily in \(\mathscr{T}_{\alpha} \cup \mathscr{T}_{\beta}\), and so the second condition for a topology is not satisfied.

(b) The unique largest topology contained in all \(\mathscr{T}_{\alpha}\) would be \(\bigcap \mathscr{T}_{\alpha}\), since it is by definition the largest set which is contained in every \(\mathscr{T}_{\alpha}\), and it is itself a topology, as proved above.

The unique smallest topology containing all \(\mathscr{T}_{\alpha}\), which we shall call \(T\), is the topology generated by the subbasis \(\bigcup \mathscr{T}_{\alpha}\).

By definition, \(T\) would contain every \(x \in \mathscr{T}_{\alpha} \forall \alpha\), every possible finite intersection of \(\bigcup \mathscr{T}_{\alpha}\), and every possible union of the finite intersections.

Now we are left to prove that this is in fact the smallest topology containing every \(\mathscr{T}_{\alpha}\).

Suppose \( T'\) is a topology that is smaller than \(T\) and contains all of \(\{\mathscr{T}_{\alpha}\}\). Then there must be elements of \(T\) which are not in \(T'\). If \(T'\) does not contain any \(x \in \mathscr{T}_{\alpha}\) for any \(\alpha\), then it does not contain all of \(\{\mathscr{T}_{\alpha}\}\). If it does not contain some finite intersection of the elements of \(\bigcup \mathscr{T}_{\alpha}\), then it is not a topology by definition. If it does not contain some union of the finite intersections of \(\bigcup \mathscr{T}_{\alpha}\), then it is not a topology by definition. Thus, \(T\) must be the smallest topology which contains all of \(\{\mathscr{T}_{\alpha}\}\).