# Maximal connected subtopologies

This is related to an older question.

Let $$(X,\tau)$$ be a topological space. Trivially, the indiscrete topology $$\{\emptyset, X\}$$ is a connected subtopology of $$\tau$$.

Is there a connected topology $$\tau_c\subseteq \tau$$ on $$X$$ such that whenever $$\tau_c'$$ is a connected topology on $$X$$ such that $$\tau_c\subseteq\tau_c'\subseteq \tau$$ then $$\tau_c=\tau_c'$$?

• The answer I gave below is incorrect; as pointed out by a comment to it, the union of a chain of topologies is not necessarily a topology. I think you should un-accept it. – user44191 Dec 3 '18 at 19:15
• Thanks for putting the effort into it... I also missed the fact that the union of a chain of topologies need not be a topology! Can you delete your answer and post as a comment, why a straightforward application of Zorn's lemma doesn't work? – Dominic van der Zypen Dec 3 '18 at 20:27
• The straightforward application of Zorn's lemma doesn't work because one of the conditions on topologies uses infinitely many sets (in fact, arbitrarily many). An example: let $X = \mathbb{N}, \tau_n$ be the topology that doesn't distinguish between $m \geq n$, and otherwise is discrete. Then $\cup \tau_n$ is the set of "eventually constant" sets (and even is countable). But it does include each of the point-sets, which contradicts the "arbitrary unions" condition on topologies. Therefore, Zorn's lemma can't be directly applied. – user44191 Dec 3 '18 at 20:46
• @ForeverMozart -- works fine if $\tau$ is connected, but if it isn't? – Dominic van der Zypen Dec 4 '18 at 7:56
• Split the real line into countably many disjoint dense sets $A_n$, e.g., $A_n=\mathbb{Q}+n\pi$ for $n\ge1$ and $A_0=\mathbb{R}\setminus\bigcup_{n\ge1}A_n$. Let $\tau_n$ be the topology generated by the usual topology together with $\{A_i:i\le n\}$. Each topology is connected, yet the union generates a topology with a pairwise disjoint cover by open sets. – KP Hart Dec 5 '18 at 8:32