Extending hyperconnected spaces 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?
 A: Consider the hyperconnected space $(X, \tau).$
The poset $P=\{\tau': \tau\subseteq \tau', (X, \tau')$ is a hyperconnected topological space$ \}$ ordered by inclusin satisfies the requirement of the Zorn's lemma (any increasing chain has an upper bound, namely topology generated by the union of the elements of the chain), so it has a maximal element, call it $\tau'.$ Then $(X, \tau')$ is a maximally hyperconnected space.
A: 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.
