A hyperconnected space is a topological space such that every two non-empty open sets have non-empty intersection. Let's call a space $(X,\cal{T})$ maximally hyperconnected if it is hyperconnected and for every topology $\cal{T'}$ with $\cal{T'} \supseteq \cal{T}$ and $\cal{T}\neq\cal{T'}$ the space $(X,\cal{T'})$ is no longer hyperconnected.


Is every hyperconnected topology contained in a maximally hyperconnected topology?



Extending hyperconnected spaces
There is a positive answer involving ultrafilters. Let $(X,\mathcal{T})$ be hyperconnected. Then note that $\mathcal{F}:=\{V\subseteq X: V\supseteq U \text{ for some non-empty } U\in\mathcal{T}\}$ is a filter. So by Zorn's Lemma, $\cal{F}$ is contained in an ultrafilter $\cal{U}$.


**Claim 1**: $(X,(\mathcal{U}\cup\{\emptyset\}))$ is a topological space with finer topology than $\mathcal{T}$.


This is easy to check.


**Claim 2**: $(X,(\mathcal{U}\cup\{\emptyset\}))$ is maximally hyperconnected.


Since $\mathcal{U}$ is a filter, every two members intersect, so it is hyperconnected. Now take any topology $\sigma$ with $\sigma\supseteq \mathcal{U}$ and $\sigma$ contains some non-empty $A\notin U$. If $\sigma$ were hyperconnected, then $\mathcal{G}:=\{V\subseteq X: V\supseteq U \text{ for some non-empty } U\in\sigma\}$ would be a filter properly containing $\mathcal{U}$, contradicting the maximality of $\mathcal{U}$. So $(X,(\mathcal{U}\cup\{\emptyset\}))$ is maximally hyperconnected.


