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I'd appreciate if someone checked whether my proofs are correct. They're not detailed enough, but I'm mainly interested whether the general approach is correct.

Let $X$ be a set, and $A$ some subset of $\mathcal P(X)$. The topology generated by $A$ is the intersection of all topologies on $X$ that contain $A$, and we will use the symbol $g(A)$ for this topology.

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1 Let $(X,\tau)$ be a topological space, with a sub-basis $\beta$. Then $g(\beta)=\tau$, because $\beta \subseteq \tau$, and $g(\tau)=\tau$, so $g(\beta) \subseteq \tau$, and $\tau \subseteq g(\beta)$, as $g(\beta)$ contains unions of finite intersections, which contain $\tau$.

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Is a slightly modified opposite implication also true?

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2 Let $g(\beta)=\tau$, then $\gamma = \beta \cup \{X\}$ forms a sub-basis for $\tau$. Proof attempt:

Let $s(\gamma)$ denote the set of unions of finite intersections of sets in $\gamma$. We want to show that $s(\gamma) = \tau = g(\beta)$. $s(\gamma) \subseteq \tau$ follows from $\gamma \subseteq \tau$. To show that $\tau \subseteq s(\gamma)$, we will show that $s(\gamma)$ is a topology - if that's true, we get that $g(\beta) \subseteq s(\gamma)$ as $s(\gamma)$ is a topology that contains $\beta$, and so $\tau \subseteq s(\gamma)$ follows.

Clearly the empty set and $X$ are both contained in $s(\gamma)$. Union of elements in $s(\gamma)$ is still in $s(\gamma)$. To show that the intersection of two elements in $s(\gamma)$ is still in $s(\gamma)$, we use the equality of $A \cap (\cup_i B_i) = \cup_i (A \cap B_i)$ twice, to show that such an element is still a union of finite intersections from $\gamma$. Therefore $s(\gamma)$ is a topology.

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Edit, as this just occured to me and it seems like a nicer way to look at it:

Alternative proofs for 2 -

We can show directly that for any $\gamma$ defined as above, $\gamma$ is a a sub-basis for exactly one topology. The topology that $\gamma$ generates has to be the same, and so we're done.

We can show that for any $\beta \subset \mathcal P(X)$, $\gamma$ defined as above is a sub-basis for some topology - $s(\gamma)$ satisfies that. Because $\beta \mapsto g(\beta)$ is a map, we see that $\gamma$ has to be the sub-basis of $g(\beta)$, as otherwise we would get a contradiction with $\beta \mapsto g(\beta)$ being a map.

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John P
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  • If I am not mistaken, ${\emptyset,A,X}$ is the smallest topology containing $A$. It is a topology, and is contained in any topology that contains $A$. – Charlie Frohman Jul 08 '17 at 23:20
  • @CharlieFrohman I think you might be replying to a previous version of the question, where I mistakenly wrote $A \subseteq X$, while it should have been $A \subseteq \mathcal P(X)$. – John P Jul 08 '17 at 23:37

1 Answers1

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There is an idea (Munkres promotes it) that a subbase for a topology $\tau$ should have $X$ as the union of its elements. This is not what a lot of texts, and me inluded, use as the definition of the term.

$\beta$ is a subbase for $\tau$ iff $g(\beta) = \tau$ (in your notation, I sometimes use $\langle \beta \rangle$ for the topology generated by $\beta$) the and a proposition says that a base for $\tau$ is given by the set of all intersections of finite subfamilies from $\beta$, including (!) the intersection of the empty subfamily, which equals $X$.

I wrote a bit more about it here

Henno Brandsma
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  • I was using that definition - a sub-basis of some topology $\tau$ is a subset of $\mathcal P(X)$, such that I can get any open set by a union of finite intersections from the subbase. I wasn't presuming that an intersection of the empty set is $X$, so I added it, hence I was using $\gamma$.

    Either way, looks like the argument in your answer you linked is pretty similar, though I didn't write out the reasons for $(\mathcal{S}^{\cap, <\infty})^{\cup}$ being a topology as rigorously.

    Thanks.

     – John P Jul 09 '17 at 07:15