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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?

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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.

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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.

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  • $\begingroup$ Is it obvious that the union of an increasing chain of hyperconnected topologies is a topology (in particular, closed under infinite unions)? $\endgroup$ Apr 29, 2015 at 17:14
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    $\begingroup$ You are right, but we can take the topology generated by the union and the resulting space is still hyperconnected. I fixed the gap. $\endgroup$ Apr 30, 2015 at 2:53

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