# Is any generator for a topology a subbase for the generated topology?

Let $$X$$ be a set, and $$S \subset P(X)$$. Let $$T(S)$$ be the topology generated by $$S$$, i.e. the smallest topology containing $$S$$. Is $$S$$ a subbase for $$T(S)$$?

Definition of subbase: $$B$$ is a subbase if the collection of open subsets consisting of all finite intersections of elements of $$B$$ together with $$X$$ forms a basis.

It does not seem obvious to me. Let $$U \in T(S)$$ be an arbitrary open. If $$S$$ is indeed a subbase, then $$U$$ is a union of (finite) intersections of elements in $$S$$. I.e. we only need one iterarion of taking intersections and unions of $$S$$.

But to form $$T(S)$$ we need at least infinite iterations of taking (finite) intersections and unions of $$S$$.

Note: there seemed to be two definitions of subbase: subbase. With the one definition the question is true obviuous, by the other it is not.

• Yes, then $S$ is a subbase of $T(S)$. – drhab Oct 31 '18 at 10:07
• Yes, this is true. But what have you tried for prove it? – Rodrigo Dias Oct 31 '18 at 10:39
• @rldias $S$ is by definition a subbase of the topology generated by $S$, so there is nothing to prove. – drhab Oct 31 '18 at 11:04
• @drhab I was using a different definition of Subbase, which I added above. – Jens Wagemaker Oct 31 '18 at 11:33
• I added the definition in the question; it is the second definition from wikipedia. So ultimately the question boiles down to establishing the equivalence of the two definitions. – Jens Wagemaker Oct 31 '18 at 11:38

If $$X$$ is a set and $$\mathcal S\subseteq\wp(X)$$ then it always generates a topology in the sense that a smallest topology that contains $$\mathcal S$$ as subcollection exists.

This merely because the intersection of all topologies on $$X$$ that contain $$\mathcal S$$ is again a topology that contains $$\mathcal S$$.

In that situation $$\mathcal S$$ is by definition a subbase of the generated topology $$\tau(\mathcal S)$$.

So a proof that $$S$$ is a subbase of $$\tau(\mathcal S)$$ is actually not needed.

What can be proved is that in this situation the topology $$\tau(\mathcal S)$$ can be described as the collection of sets that can be written as a union of finite intersections of $$\mathcal S$$.

For that it is enough to show that the latest collection is indeed a topology that contains $$\mathcal S$$ and secondly that every topology that contains $$\mathcal S$$ as a subcollection will also contain this collection as a subcollection.

edit:

Let $$\tau(\mathcal S)$$ be the smallest topology that contains $$\mathcal S$$ as a subcollection, and let $$\mathcal B$$ denote the collection of finite intersections of elements of $$\mathcal B$$. Then it is obvious that $$\mathcal B$$ is closed under finite intersection. Further it contains the empty intersection which is by convention the set $$X$$, so $$\mathcal B$$ covers $$X$$. These two conditions are exactly what is needed to ensure that the set of all unions of subsets of $$\mathcal B$$ is a topology on $$X$$, and $$\mathcal B$$ will serve as a base for that topology. The sets of this topology will be the unions of elements of $$\mathcal B$$.

Now since $$\tau(\mathcal S)$$ is a topology that contains $$\mathcal S$$ it will also contain $$\mathcal B$$ as a subcollection and will also contain the unions of sets that are in $$\mathcal B$$ as elements. Then as smallest topology that contains $$\mathcal S$$ it must coincide with that topology.

As said the collection of finite intersections of $$S$$ serves as a base for it, so also according to the alternative defintion $$\mathcal S$$ is a subbase for $$\tau(\mathcal S)$$.