Let \(f\in\mathbb C[z]\) and \(a\in\mathbb C\). Prove that the forward orbit \[ \{a,f(a),f(f(a)),\ldots\} \] is not dense in \(\mathbb C\).
Solution
Write \[ z_n\coloneqq f^{\circ n}(a) \qquad(n\ge0), \] and let \(S\coloneqq\{z_n:n\ge0\}\).
1. If \(S\) is bounded, then \(S\) is not dense in \(\CC\).
Choose \(R>0\) with \(S\subseteq\{z:\abs{z}\le R\}\). Then the nonempty open set \[ \{z:\abs{z}>R\} \] is disjoint from \(S\), so \(S\) is not dense.
2. If \(f\) is constant, then \(S\) is not dense in \(\CC\).
If \(f(z)=c\) for every \(z\), then \[ S\subseteq\{a,c\}, \] so \(S\) is finite and hence is not dense in \(\CC\).
3. If \(\deg f=1\), then \(S\) is not dense in \(\CC\).
Write \(f(z)=sz+t\) with \(s\ne0\). If \(s=1\), then \[ z_n=a+nt, \] so \(S\) lies in the affine real line \(a+\RR t\) and is not dense in \(\CC\).
Suppose \(s\ne1\). The point \[ c\coloneqq\frac{t}{1-s} \] is fixed by \(f\), and induction gives \[ z_n-c=s^n(a-c). \] If \(S\) is bounded, step 1 applies. If \(S\) is unbounded, then \(a\ne c\) and \(\abs{s}>1\). Consequently \[ \abs{z_n-c}=\abs{s}^n\abs{a-c}\ge\abs{a-c} \] for every \(n\ge0\). Thus the open disk \[ \left\{z:\abs{z-c}<\frac{\abs{a-c}}2\right\} \] is disjoint from \(S\), proving that \(S\) is not dense.
4. If \(\deg f\ge2\), then \(S\) is not dense in \(\CC\).
If \(S\) is bounded, step 1 applies. Assume therefore that \(S\) is unbounded. Since \[ \frac{\abs{f(z)}}{\abs{z}}\longrightarrow\infty \qquad\text{as }\abs{z}\longrightarrow\infty, \] there exists \(R>0\) such that \[ \abs{z}>R \quad\Longrightarrow\quad \abs{f(z)}>\abs{z}. \] Unboundedness gives an index \(n_0\) with \(\abs{z_{n_0}}>R\). Inductively, \[ \abs{z_n}>R \] for every \(n\ge n_0\). Hence \[ S\cap\{z:\abs{z}<R\} \] is contained in the finite set \(\{z_0,\ldots,z_{n_0-1}\}\).
Choose a point \(w\) in the disk \(\{z:\abs{z}<R\}\) different from each of those finitely many points. A sufficiently small open disk about \(w\) is still contained in \(\{z:\abs{z}<R\}\) and avoids \(z_0,\ldots,z_{n_0-1}\). It also avoids every \(z_n\) with \(n\ge n_0\). Thus this open disk is disjoint from \(S\), so \(S\) is not dense.
5. Q.E.D.
Step 2 handles constant polynomials, step 3 handles degree \(1\), and step 4 handles degree at least \(2\). These cases exhaust all polynomials.