Problem 4:

• (a) If $$\{\mathscr{T}_{\alpha}\}$$ is a family of topologies on $$X$$, show that $$\bigcap \mathscr{T}_{\alpha}$$ is a topology on $$X$$. Is the same true for $$\bigcup \mathscr{T}_{\alpha}$$?
• (b) Let $$\{\mathscr{T}_{\alpha}\}$$ be a family of topologies on $$X$$. Show that there is a unique smallest topology on $$X$$ containing all collections $$\mathscr{T}_{\alpha}$$, and a unique largest topology contained in all $$\mathscr{T}_{\alpha}$$.

(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 $$\{\mathscr{T}_{\alpha}\}$$.