As is well-known, the Brouwer fixed point theorem states that any continuous map from the unit disk in {\mathbb{R}^n} to itself has a fixed point. The standard proof uses the computation of the singular homology groups of spheres. The proof fails, and indeed this is no longer true, for more general compact spaces. However, the following result shows that there is a form of “approximate periodicity” that one can deduce using only elementary facts from general topology.

Consider a homeomorphism {T: X \rightarrow X} for {X} a compact metric space, i.e. a discrete dynamical system. We will prove:

Theorem 1 (Birkhoff Recurrence Theorem) There exists {x \in X} and a sequence {n_i \rightarrow \infty} with {T^{n_i} x \rightarrow x} as {i \rightarrow \infty}.

 

More can actually be said; I’ll return to this topic in the future. One doesn’t need {T} to be a homeomorphism.

Before we prove this, we need an auxiliary notion. Say that a homeomorphism {T: X \rightarrow X} is minimal if for every {x \in X}, {T^{\mathbb{Z}} x } is dense in {X}.

I claim that {T} is minimal iff there is no proper closed {E \subset X} with {TE = E} (such {E} are called {T}-invariant). This is straightforward. Indeed, if {T} is not minimal, we can take {E = \overline{ T^{\mathbb{Z}} x}}. If there is such a {T}-invariant {E}, then {T^{\mathbb{Z}} e} for {e \in E} is not dense in {X}.

Lemma 2 Let {T: X \rightarrow X} be a homeomorphism, {X} compact. Then there is a {T}-invariant {E \subset X} such that {T|_E: E \rightarrow E} is minimal.

 

Since {T} is a bijection, the intersection of {T}-invariant subsets is still {T}-invariant. Partially order the collection of nonempty {T}-invariant subsets by the relation {E \leq F} if {E \supset F}. Any chain has an upper bound then because {X} is compact and by the previous remark, so the lemma follows at once from Zorn’s lemma.

We can now prove Birkhoff’s recurrence theorem. Choose {E \subset X} to be {T}-invariant and with {T|_E} minimal. Then any point {x} of {E} will do, I claim. This follows from the next lemma.

Lemma 3 Let {T: E \rightarrow E} be minimal, {E} compact. Then for any {x \in E}, {E= \overline{T^{\mathbb{N}} x}}.

 

I claim that for any {U \subset E} open, we have

\displaystyle E = \bigcap_{\mathbb{N}} T^{-i} U.

Indeed, we have {E = \bigcap_{\mathbb{Z}} T^{-i} U} (for the complement is closed and {T}-invariant), and by compactness one can write for some {N}, {E = \bigcap_{-N}^N T^{-i}(U)}. Apply {T^{-N-1}} to both sides now.

As a result, it follows that if {U} is arbitrary, then {T^k x \in U} for some {k}, which proves the lemma. And thus the Birkhoff theorem.

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