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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.

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  • $\begingroup$ Yes, then $S$ is a subbase of $T(S)$. $\endgroup$ – drhab Oct 31 '18 at 10:07
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    $\begingroup$ Yes, this is true. But what have you tried for prove it? $\endgroup$ – Rodrigo Dias Oct 31 '18 at 10:39
  • $\begingroup$ @rldias $S$ is by definition a subbase of the topology generated by $S$, so there is nothing to prove. $\endgroup$ – drhab Oct 31 '18 at 11:04
  • $\begingroup$ @drhab I was using a different definition of Subbase, which I added above. $\endgroup$ – Jens Wagemaker Oct 31 '18 at 11:33
  • $\begingroup$ 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. $\endgroup$ – Jens Wagemaker Oct 31 '18 at 11:38
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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 $\tau(\mathcal S)$.

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