# Splitting of ordinals of oscillation ranks of a Baire $1$ function

Denny and Tang proved that

Theorem $2.3$ Let $(f_n)$ be a sequence in $\mathfrak{B}_1(K)$ converging pointwise to a function $f.$ Suppose $\sup\{ \beta(f_n):n\in\mathbb{N} \} \leq \beta_0$ and $\gamma((f_n)) \leq \gamma_0.$ Then $f$ is Baire-$1$ and $\beta(f) \leq \beta_0 \cdot \gamma_0.$

They asked problem $4.7$ (still open) in the same paper.

Open Problem: Is it true that if $f\in\mathfrak{B}_1(K)$ with $\beta(f)\leq\beta_0 \cdot \gamma_0$ for some countable ordinals $\beta_0$ and $\gamma_0.$ then there exists a sequence $(f_n)$ converging pointwise to $f$ so that $\sup_n \beta(f_n) \leq \beta_0$ and $\gamma((f_n)) \leq \gamma_0?$

They showed that the problem has positive answer if $\beta_0 = 1$ (Theorem $5.5$) or $\gamma_0$ is a limit ordinal (Theorem $4.5$).

Question: What are the known techniques to construct real-valued functions that satisfy $\sup_n\beta(f_n) \leq \beta_0$ and $\gamma((f_n)) \leq \gamma_0?$

Background and Definitions: Let $K$ be a compact metric space. We say that $f:K\rightarrow \mathbb{R}$ is of Baire $1$ if there exists a sequence of continuous functions $(f_n)$ converging to $f$ pointwise. The set $\mathfrak{B}_1(K)$ contains all real-valued Baire $1$ functions $f$ on $K$, both bounded and unbounded.

For a Baire $1$ function $f:K\rightarrow \mathbb{R}$, its oscillation rank $\beta$ is defined as follows:

For any $\varepsilon>0,$ let $K^0(f,\varepsilon) = K.$ If $K^\alpha(f,\varepsilon)$ is defined for some countable ordinal $\alpha,$ let $K^{\alpha+1}(f,\varepsilon)$ be the set of all those $x\in K^\alpha(f,\varepsilon)$ such that for every open set $U$ containing $x,$ there are two points $x_1$ and $x_2$ in $U \cap K^\alpha(f,\varepsilon)$ with $|f(x_1) - f(x_2)| \geq \varepsilon.$ For a countable limit ordinal $\alpha,$ we let $$K^\alpha(f,\varepsilon) = \bigcap_{\alpha^{\prime}<\alpha}K^{\alpha^{\prime}}(f,\varepsilon).$$ The index $\beta_K(f,\varepsilon)$ is taken to be the least $\alpha$ with $K^\alpha(f,\varepsilon) = \emptyset$ if such an $\alpha$ exists, and $\omega_1$ otherwise. The oscillation rank of $f$ is $$\beta_K(f) = \sup\{ \beta_K(f,\varepsilon):\varepsilon>0 \}.$$

If $(f_n)$ is a sequence of real-valued functions on $K,$ let $K^0((f_n),\varepsilon) = K$ for any $\varepsilon>0.$ If $K^\alpha((f_n),\varepsilon)$ has been defined for some countable ordinal $\alpha,$ let $K^{\alpha+1}((f_n),\varepsilon)$ be the set of all those $x\in K^\alpha((f_n),\varepsilon)$ such that for every open set $U$ containing $x$ and any $m\in\mathbb{N},$ there are two integers $n_1,n_2$ with $n_1>n_2 > m$ and $x' \in U \cap K^\alpha((f_n),\varepsilon)$ such that $|f_{n_1}(x') - f_{n_2}(x')| \geq \varepsilon.$ Define $$K^\alpha((f_n),\varepsilon) = \bigcap _{\alpha' < \alpha}K^{\alpha'}((f_n),\varepsilon)$$ if $\alpha$ is a countable limit ordinal. Let $\gamma_K((f_n),\varepsilon)$ be the least $\alpha$ with $K^\alpha((f_n),\varepsilon) = \emptyset$ if such $\alpha$ exists, and $\omega_1$ otherwise. The convergence index of $f$ is the ordinal $$\gamma_K((f_n)) = \sup\{ \gamma_K((f_n),\varepsilon): \varepsilon>0 \}.$$