# Existence of minimal sub-systems

A topological dynamical system is a topological space $X$ together with a continuous function $f : \ X \to X$. In the following, I will assume that $X$ is compact and Hausdorff (in other words, I work with compact topological dynamical systems).

Let $(X, f)$ be a compact topological dynamical system. A subset $A \subset X$ is said to induce a sub-system if it is non-empty, closed and $f$-invariant. Then $(A,f)$ is called a sub-system.

A sub-system $(A,f)$ is called minimal is it has no proper sub-system.

Now, a theorem by Birkhoff asserts that any compact topological dynamical system admits a minimal sub-system. The short and easy proof goes like this:

Proof

Let $\mathcal{M}$ be the set of sub-systems of $(X,f)$. This set is non-empty, as it contains $(X,f)$. The set $\mathcal{M}$ is ordered by inclusion: we write that $(A,f) \subset (A',f)$ whenever $A \subset A'$. Finally, any totally ordered subset $\{(A_i, f)\}$ of $\mathcal{M}$ has a minimal element, namely, $(\bigcap_i A_i,f)$ (which is non-empty, as a decreasing intersection of non-empty compact spaces is non-empty, closed, and $f$-invariant).

By Zorn's Lemma, $\mathcal{M}$ has minimal elements, which are minimal systems $\square$

This proof is somewhat unpalatable for me, as it uses the Axiom of Choice (which I find unintuitive when used in topology). I've seen mentioned that there exists a more technical proof which does not use the Axiom of Choice (but probably still uses the Axiom of Dependent Choice). Where may I find it (or, if the proof is short enough, what does it look like)?

• What is a dynamical system? – Asaf Karagila Nov 21 '14 at 14:23
• @Asaf Karagila: to my knowledge, there is no hard definition (the most general I can think of would be "action of a monoid", but most of the time it is given by a transformation or a flow on some space, so are actions of $\mathbb{N}$, $\mathbb{Z}$, $\mathbb{R}_+$ or $\mathbb{R}$). I don't see how it is relevant to the question at hand, though. – D. Thomine Nov 21 '14 at 14:52
• You find the axiom of choice unintuitive in topology? Does that mean that you don't like to use Tychonoff's theorem? – Asaf Karagila Nov 21 '14 at 14:52
• Also, my first question is very relevant, since I know a thing or two about the axiom of choice, but nothing at all about a dynamical system. If I knew what is the definition of a dynamical system (that you're using here), I could have tried to come up with a counterexample or a proof. So my question is relevant for me. – Asaf Karagila Nov 21 '14 at 14:53
• @Asaf Karagila: I don't. For me, things work pretty nicely as long as you only use countable products (e.g. separable Banach spaces behave nicely, but $\mathbb{L}^\infty$ is nastier). That's only a matter of taste and motivation. – D. Thomine Nov 21 '14 at 14:54