IMC 2002 · Day 1 · P4

If for some $n > m$ the equality $p_m = p_n$ holds then $T_p$ is a finite set. Thus we can assume that all points $p_0, p_1, \dots$ are distinct. There is a convergent subsequence $p_{n_k}$ and its limit $q$ is in $T_p$. Since $f$ is continuous $p_{n_k + 1} = f(p_{n_k}) \to f(q)$, so all, except for finitely many, points $p_n$ are accumulation points of $T_p$. Hence we may assume that all of them are accumulation points of $T_p$. Let $d = \sup \{ |p_m - p_n| : m, n \ge 0 \}$. Let $\delta_n$ be positive numbers such that $\sum\limits_{n=0}^{\infty} \delta_n < \frac{d}{2}$. Let $I_n$ be an interval of length less than $\delta_n$ centered at $p_n$ such that there are there are infinitely many $k$'s such that $p_k \notin \bigcup\limits_{j=0}^{n} I_j$, this can be done by induction. Let $n_0 = 0$ and $n_{m+1}$ be the smallest integer $k > n_m$ such that $p_k \notin \bigcup\limits_{j=0}^{n_m} I_j$. Since $T_p$ is closed the limit of the subsequence $(p_{n_m})$ must be in $T_p$ but it is impossible because of the definition of $I_n$'s, of course if the sequence $(p_{n_m})$ is not convergent we may replace it with its convergent subsequence. The proof is finished.

Remark. If $T_p = \{p_1, p_2, \dots\}$ and each $p_n$ is an accumulation point of $T_p$, then $T_p$ is the countable union of nowhere dense sets (i.e. the single-element sets $\{p_n\}$). If $T_p$ is closed then this contradicts the Baire Category Theorem.