2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $\ \Gamma$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Wow! Enumerating the square like that is a really cool example! $\endgroup$ – Daron Apr 8 '18 at 17:12
  • 1
    $\begingroup$ 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
1
$\begingroup$

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).

$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.