Continuum $=$ compact connected metric space.

Let $X$ be a continuum. $X$ is indecomposable means that every proper subcontinuum of $X$ is nowhere dense in $X$.

It is easy to see that if $X$ is indecomposable then every connected open subset of $X$ is dense in $X$.

Question. Are these two conditions equivalent?

Given the wealth of examples in continuum theory, the answer is likely no. So what is an example of a decomposable continuum all of whose connected open subsets are dense?

EDIT: I have constructed an example; the two conditions are NOT equivalent. Before revealing it, I will leave the bounty open in hopes of attracting more examples.

  • $\begingroup$ So, what is your counterexample? $\endgroup$ Aug 26, 2018 at 14:06

1 Answer 1


**Edit 1.a:**This construction tried to provide a counterexample, but is wrong. I give more details below.

Glue two solenoids as follows.

Let $K$ be a Cantor set (e.g. iterate $[0,1]\to [0,\frac 1 3] \cup [\frac 2 3,1]$). Let $D\subset K$ be a closed nowhere dense Cantor subset (e.g. iterate $[0,1]\to [0,\frac 1 9] \cup [\frac 8 9, 1]$).

Let $S_1,S_2$ be two solenoids over circles $C_1,C_2$ respectively. Essentially, a solenoid is a space $S$ that fibres over a circle $S\to C$, with all the fibres homeomorphic to Cantor sets, and with a certain nontrivial twist. A solenoid is an indecomposable continuum, is homogeneous, and the only proper subcontinua are arcs.

On each circle $C_i$ consider two points $P_i^0, P_i^\pi$, and let $F_i^\alpha$ be the corresponding fibers on $S_i$, for $i=1,2$ and $\alpha=0,\pi$. Let homeomorphisms $a: F_1^0\to D$, $b:K\to F_2^0$, $c: F_2^\pi\to D$, $d:K\to F_1^\pi$ be given.

Glue $F_1^0\cup (d\circ c)(F_2^\pi)\subset S_1$ to $(b\circ a)(F_1^0)\cup F_2^\pi\subset S_2$ via the above identification, to obtain the required space $X$.

$X$ is clearly a continuum, and is decomposable, because is the union of two proper subcontinua (copies of $S_1,S_2$).

Edit 1.b: The observation motivating the example was that $S_1$ and $S_2$ are separately indecomposable, and an open set $U$ meeting $S_1$ necessarily meets the fiber $F_1^0$ and since $F_1^0$ is nowhere dense in $F_2^0$ it follows that $U$ meets $F_2^\pi$ as well. This property is achieved in a less trivial way than just identifying (an open subset of) a fiber of $S_1$ with a similar subset of $S_2$. Indeed there is no small neighborhood of $F_2^0$ retracting on it. However this does not really help in solving the question.

  • $\begingroup$ Okay I see the construction, but I'm still having trouble with the last paragraph. Are you saying that $U\cap S_1$ must be connected? $\endgroup$ May 30, 2017 at 18:40
  • $\begingroup$ I was not saying this; I was thinking of something more involved. But I'm wrong nonetheless. $\endgroup$ May 31, 2017 at 1:59
  • $\begingroup$ My idea is to glue only two indecomposable continua S1,S2 such that (1) a connected open set U necessarily meets the intersection F; (2) the space X is not "locally connected to" F; (3) this forces U to be dense in S1 and S2 separately. Is it possible to achieve this? The true reason for the failure of (3) in the construction above, is that the solenoid is not crooked enough. The next obvious trial is to glue two pseudoarcs in their "middle" (i.e. if a pseudoarc is in the unit square, the middle is the intersection with the line y=0.5). $\endgroup$ May 31, 2017 at 13:49

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