Let X be an infinite regular topological space which is connected and locally connected.

Question. 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.

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    $\begingroup$ 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 $\endgroup$ Jun 3, 2014 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, 2014 at 20:08
  • $\begingroup$ U is open iff for each point p of U there exist a (non-degenerate) connected subset V of U containing p, such that the complement of V (in the space X) is a closed connected subset of X. $\endgroup$ Jun 5, 2014 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. $\endgroup$ Jun 5, 2014 at 15:07
  • $\begingroup$ Maybe useful reference: ncatlab.org/nlab/show/connectology $\endgroup$ Aug 15, 2020 at 6:16

1 Answer 1


The answer is no. We can construct a connected and locally connected space $X$ in the plane such that:

  1. $X$ has no cut points; and
  2. there is a point $p\in X$ such that $X\setminus U$ is disconnected for every sufficiently small open subset $U$ of $X$ containing $p$.

Let $p=\langle 0,0\rangle$ and let $\{S_n:n<\omega\}$ be a collection of simple closed curves in the plane such that $S_n\cap S_m=\{p\}$ for all $n\neq m$, each $S_n$ intersects the circle of radius $1$ around $p$, and each $S_n\setminus \{p\}$ is open in $S:=\bigcup \{S_n:n<\omega\}$. For each $n<\omega$ connect $S_{n}$ to $S_{n+1}$ with an arc $A_n$ which has one endpoint in $S_{n}$ and the other endpoint in $S_{n+1}$, such that $A_n$ is contained in the disc of radius $1/n$ around $p$. Let $X=S\cup \bigcup \{A_n:n<\omega\}$. If $U$ is any open set containing $p$ of diameter less than $1$ then $X\setminus U$ is not connected, because for sufficiently large $n$ the curve $S_n$ reaches outside of $U$ but its connecting arcs $A_{n-1}$ and $A_n$ are contained in $U$.

The example is Polish but non-compact. I'm not sure if there is a compact counterexample.


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