Let X be an infinite regular topological space which is connected and locally connected. If no point of X is a cut point, does X always have base of connected open sets whose complements (with respect to X) are also connected? The answer is "YES" if X is a real Banach space whose dimension is at least 2. The motivation for this question is to obtain a simple characterization of as large a class of spaces as possible, which have the property that their topology is uniquely determined when the collection of their connected subsets is specified. Note that closed connected sets can be defined in terms of connected sets. They are those connected sets whose union with the singleton of any point not belonging to them is not connected. Then we can take the complements of the closed connected sets to be the base of our topology.
9

1$\begingroup$ How would you define the separation properties (e.g. Hausdorff, completely regular) in terms of connected sets? Gerhard "Not To Mention Compactness Too" Paseman, 2014.06.03 $\endgroup$ – Gerhard Paseman Jun 3 '14 at 20:01

$\begingroup$ $U$ is open iff $\forall$ connected $C, \forall x \in C \cap U, \exists$ connected $C' \subset C \cap U$ with $\{x\} \subsetneq C'$? $\endgroup$ – Matt F. Jun 3 '14 at 20:08

$\begingroup$ U is open iff for each point p of U there exist a (nondegenerate) connected subset V of U containing p, such that the complement of V (in the space X) is a closed connected subset of X. $\endgroup$ – Garabed Gulbenkian Jun 5 '14 at 14:56

$\begingroup$ I really must apologize. I just discovered that this question is almost a duplicate of question No. 126897, which I asked over a year ago, and which I had completely forgotten about because it received no responses of any sort. However the new question clearly differs from the old one in one respect. The old question contains no mention of cut points. $\endgroup$ – Garabed Gulbenkian Jun 5 '14 at 15:07