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I want to understand the following argument found in Katok's Introduction to Dynamical Systems, page 27:

Claim: If $\alpha$ is irrational, then every orbit is dense in the circle.

Proof: Let $A\subset S^1$ the closure of an orbit. If the orbit is not dense, then $S^1 \setminus A$ is a nonempty open invariant set which consist of disjoint intervals. Let $I$ be the longest of those intervals. [...]

I don't understand the bold argument above, why there exist a longest interval? As I can see, an open set in $S^1$ can be written as a numerable union if disjoint intervals, why can't happen that there is always an interval slighly larger -but bounded- so there isn't any largest interval?

HeMan
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1 Answers1

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If $S^1\setminus A$ contains infinitely many disjoint intervals, their lengths has to go to zero, since the sum of their lengths should converge.

So for any $\epsilon>0$, there's only finitely many elements of $S^1\setminus A$ with length greater than $\epsilon$, therefore has a maximal element.

Qidi
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