In my Point Set Topology class, we learned about topologies generated by bases and sub-bases. However, I think I have a notion of a topology generated by arbitrary subsets of the powerset of a given set, but I don't know whether this is a well known topology, or whether the construction is faulty.
Let $S$ be a set, and let $X\subset\mathcal P(S)$. Consider $H$, the set of all topologies on $S$ which contain $X$. Note that $H\neq\varnothing$ thanks to the discrete topology. Consider the reverse inclusion partial ordering ($J \geq K \longleftrightarrow J \subset K$) on $H$, and consider an arbitrary totally ordered subset of $H$, $H'$. Note that $X\subset\bigcap H'$, and that $\bigcap H'$ is a topology on $S$.
We can show this in 3 steps
- $\varnothing, S\in\bigcap H'$: obvious
- For all $J\subset\bigcap H'$, $\bigcup J\in\bigcap H'$: $J$ is a subset of all the topologies in $H'$, so $\bigcup J$ is in all of the topologies of $H'$, implying that $\bigcup J\in \bigcap H'$.
- For $Y,Z\in\bigcap H',\;Y\cap Z\in\bigcap H'$: Again, $Y,Z$ are in every topology in $H'$, which implies that $Y\cap Z$ is as well, so $Y\cap Z\in\bigcap H'$.
Note that $\bigcap H'\geq J$ for all $J\in H'$, and that $\bigcap H'\in H$. Hence, we can apply Zorn's Lemma to show that there exists maximal elements in $H$.
Assume for the sake of contradiction that there exist $K,L\in H$ such that $K\neq L$ and $K,L$ are both maximal. Since $K\not<L$ and $L\not< K$, we have that $K\cap L$ is a topology (this is pretty trivial to show) and that $K\cap L<K,L$, a contradiction. So, we know that we have a single maximal element. This implies that there exists a topology $X'$ such that for every topology $Y$ in $H$, $X'\subset Y$.
So, I propose defining the topology generated by the set $X$ to be $X'$. It seems to be consistent with the notions of topologies generated by bases and sub-bases, but generalizes to arbitrary sets.
Is this construction valid? Is it useful? If the answer to both of these questions is yes, where can I find it in the literature?
Best Answer
If $H$ is a non-empty family of topologies on $S$ then $\cap H$ is a topology on $S.$
If $X\subset \mathcal P(S)$ and $H$ is the set of all topologies $T$ on $S$ such that $T\supset X$ then (i) $ H\ne \emptyset$ because $\mathcal P(S)\in H,$ and (ii) $\cap H$ is the unique $\subset$-minimal member of $H.$
We do not need Zorn's Lemma. For example, to show that $\cap H$ is closed under unions: If $G\subset \cap H$ then for any $T\in H$ we have $G\subset T$ so $\cup G\in T$ (because $T$ is a topology). So $\forall T\in H\,(\cup G\in T),$ so $\cup G\in \cap_{T\in H}(T)=\cap H.$
In set theory the notation $[V]^{<\omega}$ denotes the set of all finite sub-sets of $V.$ If $V\subset \mathcal P(S)$ and $\cup V=S$ then $\{\cap W: W\in [V]^{<\omega}\}$ is a base for a topology $U$ on $S,$ and $V$ is called a sub-base for $U$. In terms of the $X$ and $H$ of your Q, if $V=X\cup \{S\}$ (to ensure that $\cup V=S$) then $U=\cap H.$
For any $X \subset \mathcal P(S), $ I see no reason not to call $\cap H$ the topology on $S$ generated by $X.$ A standard term is to call $\cap H$ the weakest topology $T$ on $S$ such that $T\supset X.$