In his well-known paper Bellamy constructs an indecomposable continua with exactly two composants. The setup is as follows:

We have an inverse-system $\{X(\alpha); f^\alpha_\beta: \beta,\alpha < \omega_1\}$ of metric indecomposable continua and retractions. For each $X(\alpha)$ a composant $C(\alpha) \subset X(\alpha)$ is specified and each $C(\alpha)$ maps into $C(\beta) $.

The inverse limit $X$ has exactly two composants. The first is the union $\bigcup\{X(\beta): \beta < \omega_1\}$ where we identify $X(\beta)$ with the set of sequences $(x_\alpha) \in X$ with $x_\beta = x_{\beta+1} = x_{\beta+2} = \ldots . $ The second composant is the inverse limit $\{C(\alpha); f^\alpha_\beta: \beta,\alpha < \omega_1\}$. Observe there is no reason a priori for the second composant to be nonempty. However I do not believe an example is know.

My question is an easier one. Can you think of an example of a metric continuum $M$ and a $\omega_1$-indexed decreasing nest of dense semicontinua with empty intersection? We call the set $S \subset M$ a semicontinuum to mean for each $x,y \in S$ there exists a continuum $K$ with $\{x,y\} \subset K \subset S$.

If the second composant was empty the family $\{f^\alpha_0(C(\alpha)): \alpha < \omega_1\}$ would be such a nest for $M = X_0$.

If we index by $\omega$ instead an example is easy to come by. Let $M$ be the unit disc and $Q = \{q_1,q_2, \ldots\}$ and enumeration of the rational points on the boundary. Let $S(n)$ be formed by drawing the straight line segment from each element of $\{q_n,q_{n+1}, \ldots\}$ to each rational point of $(0,1/n) \times \{0\}$. Then add in $(0,1/n) \times \{0\}$ itself to make the space a semicontinuum.

Indexing by $\omega_1$ must somehow get around the fact that any $\omega_1$ decreasing nest of closed subsets of a metric continuum eventually stabilizes.

It feels like this would be easier if we assume the Continuum Hypothesis.


I am not assuming any CH.

Let $\ \Omega\ := \{\alpha: \alpha<\omega_1\},\ $ and $\ i:\Omega\rightarrow [0;1]\ $ be injective and such that $\ \Gamma\ :=\ i(\Omega)\ $ is condensed in $\ [0;1].\ $ More generally, let $\ \Gamma_\alpha\ :=\ i([0;\alpha)).\ $ Also let $\ M:=[0;1]^2.\ $ Then

$$ S_\alpha\,\ :=\,\ (\Gamma\setminus\Gamma_\alpha)\times[0;1] \ \cup\ [0;1]\times(\Gamma\setminus\Gamma_\alpha) $$

for $\ \alpha<\omega_1,\ $ is the required decreasing $\omega_1$-sequence which has empty intersection.

Condensed means that every non-empty open subset of $\ [0;1]\ $ has an uncountable intersection with $\ \Gamma$.


Indeed, if CH holds then there is an enumeration $\{x_\alpha:\alpha<\omega_1\}$ of all the points in the square $M:=[0,1]^2$. Then take $S(\alpha)=M\setminus \{x_\beta:\beta<\alpha\}$.

Each $S(\alpha)$ has countable complement and is therefore path/arc-connected $-$ for every two points $x,y\in S(\alpha)$ there is actually an arc in $S(\alpha)$ consisting of two straight line segments which joins $x$ and $y$. So $S(\alpha)$ is a semicontinuum.

It is clear that $\overline{S(\alpha)}=M$, the $S(\alpha)$'s are decreasing, and $\bigcap\{S(\alpha):\alpha<\omega_1\}=\varnothing$.

Side note: I thought this problem was interesting when I first studied Bellamy's example a couple years ago. I've also wanted to use his inverse limit of retractions method to find a metrizable decomposable continuum $X$ which has only finitely many (or countably many) equivalence classes $x/\sim$, where $x\sim y$ if and only if there is a nowhere dense subcontinuum of $X$ containing $x$ and $y$. I believe this is also an open problem.

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    $\begingroup$ Wow! Enumerating the square like that is a really cool example! $\endgroup$ – Daron Apr 8 '18 at 17:12
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    $\begingroup$ Presumably you want at least two of these 'pseudo-composants' to exist? $\endgroup$ – Daron Apr 8 '18 at 17:12
  • $\begingroup$ Yes, that is correct. We'd like to know if there is some number of these eq. classes between $1$ and $2^\omega$. Of course, Bellamy provided indecomposable examples, and they can be modified to get decomposable continua with every number $\leq \omega$. You can even get decomposable with exactly two, each dense. Nothing is known in the metrizable setting other than $1$ and $2^\omega$ (which is possible by attaching two metric indecomposable continua at a point). $\endgroup$ – D.S. Lipham Apr 8 '18 at 18:47
  • $\begingroup$ Variation of this problem: Is there a continuum $X$ with a point $p$ such that (1) for every two points $x,y \in X\setminus \{p\}$ there is a nowhere dense subcontinuum of $X$ containing $x$ and $y$; and (2) every proper subcontinuum of $X$ meeting $p$ and $X\setminus \{p\}$ has interior? This is unknown, even in the general setting! I was (and still am) shocked that nobody has been able to answer that. $\endgroup$ – D.S. Lipham Apr 8 '18 at 18:53
  • $\begingroup$ Do these problems have a name in the literature or are they entirely of your own devising? $\endgroup$ – Daron Apr 8 '18 at 19:40

Let $\ \Omega\ := \{\alpha: \alpha<\omega_1\},\ $ and $\ i:\Omega\rightarrow [0;1]\ $ be injective. Let $\ M:=[0;1]^2.\ $ Then

$$ S_\alpha\ :=\ M\ \setminus \ i([0;\alpha))\times(0;1] $$

for $\ \alpha<\omega_1,\ $ is the required decreasing $\omega_1$-sequence

A VARIANT A similar example admits

$$ M:=[0;1]\times\{0\}\cup C\times[0;1]$$

This time continuum $\ M\ $ is $1$-dimensional (which feels thematic).

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    $\begingroup$ Can you modify your example so the intersection is empty? $\endgroup$ – Daron Apr 8 '18 at 17:21
  • $\begingroup$ @Daron, I have a somewhat similar example, with an empty intersection, when I assume CH. Its simplicity is comparable to David's example. The two would be different, they would have each their own small advantages (small because the whole thing is a miniature in each case). $\endgroup$ – Wlod AA Apr 9 '18 at 2:16
  • $\begingroup$ @Daron, done. See my other answer. $\endgroup$ – Wlod AA Apr 9 '18 at 19:43

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