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I have a doubt in the proof of the following theorem given in An introduction to Ergodic Theory by Peter Walters.

If $T:K \to K$ is the homeomorphism of the unit circle, then $h(T)=0.$

Here $h(T)$ is the entropy of $T.$ The proof goes as follows:

We assume length of the circle is $1$ and choose $\epsilon>0$ such that $$d(x,y)<\epsilon\implies d(T^{-1}x,T^{-1}y)<\frac 14.$$

Clearly, $$r_1(\epsilon,K)\leq \left\lfloor \frac{1}{\epsilon} \right\rfloor +1$$ where $r_n(\epsilon,K)$ denotes the minimum cardinality of $(n,\epsilon)$ spanning sets of $K.$ Let $F$ be the $(n-1,\epsilon)$ spanning set of cardinality $r_{n-1}(\epsilon,K).$ Add points to the set $T^{n-1}(F)$ so that new intervals have length less than $\epsilon.$ We have added at most $\lfloor \frac{1}{\epsilon} \rfloor+1$ points. Let $E$ denotes the set of these new points and $$F'=F \cup T^{-(n-1)}(E).$$ Claim is that $F'$ is an $(n,\epsilon)$ spanning set of $K.$ Let $x \in K.$ Then there exists $y \in F$ such that $$\max_{0\leq i\leq n-2}d(T^ix,T^iy)\leq \epsilon.$$ If $d(T^{n-1}x,T^{n-1}y)\leq \epsilon,$ then we are done. If not, then let $I$ denothe the arc with end points $T^{n-1}x$ and $T^{n-1}y$ which is mapped by $T^{-1}$ to the arc $I'$ with end points $T^{n-2}x$ and $T^{n-2}y$ and length less than or equal to $\epsilon.$ Choose $z \in F',T^{n-1}z\in I$ such that $$d(T^{n-1}x,T^{n-1}z)\leq \epsilon.$$ Then $T^{n-2}z \in I'$ and $$d(T^{n-2}x,T^{n-2}z)\leq \epsilon.$$

The interval $I'$ is mapped by $T^{-1}$ to an interval $I''$ with end points $T^{n-3}x$ and $T^{n-3}y$ and length less than $\frac 14.$

Till this point everything is fine. But I don't understand the line that follows:

Therefore, the length of $I''$ is less than or equal to $\epsilon.$

I understand the length of $I''$ is less than $\frac 14,$ but why should it mean that it is less than $\epsilon.$ Also, there is no special use of $\frac 14$ in proof. Was there any specific reason to use $\frac 14$ while applying uniform continuity?

Sahiba Arora
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1 Answers1

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What does $\max_{0\leq i\leq n-2}d(T^ix,T^iy)\leq \epsilon$ tell you about $d(T^{n-3} x,T^{n-3}y)$?

... Assuming $n \geq 3$, which it is because the choice of epsilon forces $n > 4$. That is, since $d(x,y) < \epsilon \implies d(T^{-1} x,T^{-1}y) < \frac{1}{4}$, we need at least four $\epsilon$ neighborhoods for their preimages to cover the original circle.

Eric Towers
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  • Ah. I was confused because of the "therefore." – Sahiba Arora Jan 12 '18 at 01:09
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    Yeah... The "therefore" isn't quite the best choice. Honestly, I'm a little irritated by the author. The only inference leading to a distance being $< \frac{1}{4}$ requires the image distance be $< \epsilon$, but in the subsequent argument these are only given $\leq \epsilon$. (The difference is straightforwardly corrected, but the argument is a little sloppy.) – Eric Towers Jan 12 '18 at 01:19