# General topology on Kolmogorov spaces

Let $$X$$ be a Kolmogorov space in which every non- empty open set is infinite. Show that there exists a quasi-maximal topology on $$X$$ which is finer than the given topology. (show that the set of topologies on $$X$$ in which all the non-empty open sets are infinite is inductive.)

I used that if given set is infinite, then $$X$$ has no isolated point. But how does maximality forces a finer topology than given topology?

• Please include your thoughts on the problem and attempts you have made so far. – Thomas Shelby Feb 21 at 17:04
• i used that if given set is infinite, then $X$ has no isolated point. but how does maximality forces a finer topology than given topology – prabhjotsingh Feb 21 at 17:08

Let $$\mathcal{P}$$ be the poset of all topologies on $$X$$, ordered by inclusion, that are at least as big (fine) as $$\tau_X$$ (the given topology) and that also have the property that all open sets are infinite (so that in particular those topologies have no isolated points).
Let $$\mathcal{C}$$ be a chain in $$\mathcal{P}$$.
Then let $$\tau$$ be the topology generated by $$\bigcup \mathcal{C}$$ (so take that collection of sets as its subbase). Then this topology trivially contains $$\tau_X$$ too (as all members of $$\mathcal{C}$$ do) and $$\tau$$ has no finite open set: suppose that it had, then there would be finitely $$O_1,\ldots, O_n \in \bigcup \mathcal{C}$$ such that $$O:=O_1 \cap O_2 \cap \ldots O_n$$ is finite (as finite intersections from the subbase form a base for $$\tau$$), with $$O_i \in \tau_i \in \mathcal{C}$$ and as $$\mathcal{C}$$ is a chain in our poset, there is one $$\tau_j \in \{\tau_1,\ldots,\tau_n\}$$ that contains all of the $$O_i, i=1,\ldots,O_n$$. But then $$\tau_j$$ would have already contained the finite open set $$O$$ which is not the case (as it is in $$\mathcal{P}$$). So $$\tau$$ is a member of $$\mathcal{P}$$ and an upperbound for the chain $$\mathcal{C}$$. So $$\mathcal{P}$$ is inductive.
Zorn's lemma now immediately implies that $$\mathcal{P}$$ has a maximal element $$\tau_{\text{max}}$$ and this is the required quasi-maximal topology finer than $$\tau_X$$.