Let $f$ be a polynomial of degree greater than 1. Prove that the following set is compact:

$K(f)= ( z \in \mathbb{C}: f(f(f(...f(z)...)))$ does not diverge )

Alternatively, the set of all complex numbers such that an infinite number of iterations by $f$ does not diverge ( not necessarily converge )

My attempt:

Setting this proof up I went with the “closed and bounded” approach. I believe I managed to get boundedness by appealing to the fact that after each iteration the degree of the polynomial will grow fast. However proving this set is closed has alluded me; I tried setting up a sequence that converges to a point and showing that point is in $K(f)$. While doing this I tried using the fact that each iteration is a polynomial, and hence, continuous, but this doesn’t give a straightforward explanation as to why the infinite iteration will not diverge.

Background: Complex Qual question #$7$ (www.math.tamu.edu/graduate/phd/quals/ncomplex/a17.pdf)

  • 2
    $\begingroup$ To be specific, the question you have written and the question posed are slightly different. The question in the link specifies that $f^{[n]}(z)\not\to\infty$ whereas you only require it not diverge (which seems to be to saying that the iterations do converge). $\endgroup$ – Clayton May 30 '18 at 20:50
  • $\begingroup$ Well that changes everything... do you have a reference for the notation? $\endgroup$ – Sean Nemetz May 30 '18 at 20:57
  • $\begingroup$ Sean, the notation comes from the link you provided. $\endgroup$ – Clayton May 30 '18 at 20:58
  • $\begingroup$ I mean for the notation’s meaning- not “tending” to $\infty$ does not imply it converges. $\endgroup$ – Sean Nemetz May 30 '18 at 21:04
  • $\begingroup$ I'm saying your statement that the iterations do not diverge implies the iterations do converge. The statement: $f^{[n]}(z)\not\to\infty$ does not imply the iterations converge (since your statement and the linked question imply different things, their meaning cannot be equivalent). $\endgroup$ – Clayton May 30 '18 at 21:09

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