Is there a universal $\omega$-limit set?

Question

For the purposes of this question, a dynamical system means a compact metric space $X$ together with a continuous map $f: X \to X$.

For $x \in X$, the $\omega$-limit set of $x$, denoted $\omega(x)$, is the set of limit points of the orbit of $x$. That is, $$\omega(x) = \bigcap_{n \in \omega} \overline{\{f^m(x) : m > n\}}.$$ It's easy to check that $(\omega(x),f)$ is a dynamical system (i.e., $\omega(x)$ is closed under $f$ and closed topologically). Let's say that $(X,f)$ is an abstract $\omega$-limit set if it is isomorphic to the $\omega$-limit set of some point in some dynamical system.

Is there a dynamical system $(X,f)$ such that every abstract $\omega$-limit set is a quotient of $(X,f)$?

[In this context, an isomorphism $(X,f) \rightarrow (Y,g)$ means a homeomorphism $h: X \rightarrow Y$ such that $h \circ f = g \circ h$. A quotient $(X,f) \rightarrow (Y,g)$ means a continuous surjection $h: X \rightarrow Y$ such that $h \circ f = g \circ h$.]

I suspect the answer might be negative, but I can't think of a way to get at a proof.

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Motivation: If we replace "compact metric" with "compact Hausdorff" (to broaden our notion of dynamical systems a bit), then there is a universal abstract $\omega$-limit set, namely $\mathbb{N}^* = \beta \mathbb{N} \setminus \mathbb{N}$ together with the shift map (the "shift map" being the map sending an ultrafilter $\mathcal F$ to the ultrafilter generated by $\{A+1 : A \in \mathcal F\}$). It seems to me that some smaller system should suffice to "capture" just the metric systems. Ideally, one would like a metric system to do this -- hence the question!

Potentially helpful information: The metrizable abstract $\omega$-limit sets have a very nice topological characterization due to Bowen. Namely, a metrizable system $(X,f)$ is an abstract $\omega$-limit set if and only if it is chain transitive. [$(X,f)$ is chain transitive if for some (equivalently, any) metric $d$ on $X$, for any two points $x,y \in X$, and for any $\varepsilon > 0$, there is a sequence $x = x_0, x_1, x_2, \dots, x_n = y$ such that $d(f(x_i),x_{i+1}) < \varepsilon$ for all $i < n$. The idea here is that this sequence is a "pseudo-orbit" (with error $\varepsilon$) from $x$ to $y$.]

Also possibly relevant is the well-known topological fact that every compact metric space is a continuous image of the Cantor space. So if $(X,f)$ is a system providing a positive answer to my question, I suspect $X$ should be the Cantor space.

Answer 1

The answer is no, there is no universal $\omega$-limit set.
The same is true for the classes of metric minimal dynamical systems and metric dynamical systems in general. I have written up proofs of these facts:

http://www.math.uni-hamburg.de/home/geschke/papers/NoUniversalMetricOmegaLimitSet.pdf

The basic idea of the proof is as follows: as Will Brian already conjectured, it is enough to consider dynamical systems with 0-dimensional phase spaces. Now, if $(X,f)$ is a dynamical system with a 0-dimensional metric phase space $X$ and there is a morphism from $(X,f)$ onto a dynamical system $(Y,g)$ with $Y$ 0-dimensional, then the Boolean algebra of clopen subsets of $Y$ embeds into the algebra of clopen subsets of $X$ by an embedding that respects the dynamical systems. Since $X$ is metric, its Boolean algebra of clopen subsets is countable. Using Sturmian subshifts we can cook up enough metric minimal dynamical systems $(Y,g)$ (that are automatically abstract $\omega$-limit sets) so that not all the corresponding Boolean algebras (with actions on them) embed into a single countable Boolean algebra (with action), showing that there are no metric universal dynamical systems.

The most general result is this: There is no metric dynamical system that maps onto all Sturmian subshifts. The Sturmian subshifts are minimal and hence abstract $\omega$-limit sets. Here it doesn't make a difference whether we consider $\mathbb Z$-actions, i.e., compact spaces with a homeomorphism, or $\mathbb N$-actions, i.e., compact spaces with a continuous map as in the original question.