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Let $H$ be a collection of subsets of a set $X$. Then we define the smallest topology on $X$ containing $H$ in the obvious way: $$\tau_{H}=\bigcap \sigma$$ where the intersection is taken over all topologies $\sigma$ on $X$ which contain $H$. It's not hard to prove that $\tau$ is a topology, and that it contains every other topology which contains $H$. However, this isn't a very 'direct' definition, in that it doesn't give us much to work with.

Now, let $(X,\tau)$ be a topological space. We say that a collection of open sets $N$ is a subbasis for $\tau$ if any $U \in \tau$ can be written as an arbitrary union of finite intersections of sets in $N$. Equivalently, any $x \in U$ has an open neighbourhood consisting of finite intersections of elements in $N$.

Now, it is certainly true that every element of $H$ is open in $\tau_{H}$. I also think $H$ is a subbasis for $\tau_{H}$.

For a collection $H \subset \mathcal{P}(X)$ we define the span of $H$, $\langle H \rangle$, as the (deep breath) collection of all unions of finite intersections of elements of $H$. $\langle H \rangle$ is closed under finite intersections and arbitrary unions, and contains the empty set as the trivial intersection. It's not clear to me that $\langle H\rangle$ contains $X$, but that seems sensible because $H$ is just any old collection of subsets (we could always embed it in a larger set).

I think, then, that $\langle H \cup \{X\} \rangle$ is a topology on $X$ containing $H$, and it follows that it equals $\tau_{H}$. Is this true?

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    $\begingroup$ The set $X$ is the trivial intersection, the empty set is the empty union. Therefore $\langle H \rangle$ already contains $X$ and it's in fact true that if $H$ is a subbasis for a topology $\tau_H$, then $\tau_H = \langle H \rangle.$ $\endgroup$ – sqtrat Jun 3 '16 at 10:18
  • $\begingroup$ Have a look at this answer $\endgroup$ – drhab Jun 3 '16 at 10:21
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Indeed. $X$ is the intersection of the empty subcollection of $H$.

In general, when we define the set $H^\cap$ of all finite intersections of $H$, then we mean all sets of the form $\bigcap H'$ where $H' \subseteq H$ is finite.

$\bigcap H' = \{x \in X: \forall A \in H': x \in A \}$.

This means that we can take, if $A \in H$, $H' = \{A\}$ and get $\bigcap H' = A$, so $H \subseteq H^\cap$. Also we get all sets of the form $A_1 \cap A_2$, where $A_1,A_2 \in H$, as $\bigcap \{A_1,A_2\}$, etc.

But we can also take $\bigcap \emptyset = X$, because the $\forall$ condition is trivially true. So $X \in H^\cap$ as well. Note that $\emptyset \subseteq H$ is quite finite.

So then we can show that $H^\cap$ is closed under finite intersections itself, as $\bigcap H_1 \cap \bigcap H_2 = \bigcap (H_1 \cup H_2)$, where $H_1 \cup H_2$ is finite whenever $H_1$ and $H_2$ are.

So $H^\cap$ is a base for a topology $\mathcal{T}$ (closed under intersections and union is $X$ implies this), which contains $H$, so $\mathcal{T}_H \subseteq \mathcal{T}$, but as any topology is closed under finite intersections and unions, if $\mathcal{T}'$ contains $H$, it contains $H^\cap$ so also $\mathcal{T}$, so $\mathcal{T}_H = \bigcap \{\sigma: \sigma \text{ a topology and } H \subseteq \sigma \} \subseteq \mathcal{T}$, and we equality.

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