# Example of a Baire Class $1$ function $f$ satisfying $\omega\cdot n<\beta(f)\leq \omega\cdot (n+1)$ for some natural number $n\geq 1.$

Definitions: Let $X$ be a Polish space (separable completely metrizable topological space). A function $f:X\to\mathbb{R}$ is Baire Class $1$ if it is a pointwlise limit of a sequence of continuous functions.

Baire proved that $f$ is Baire Class $1$ if and only if for every nonempty closed set $P\subseteq X,$ there exists a point of continuity of $f$ in $P.$ To measure the 'quantitative discontinuities' of $f,$ the oscillation rank is introduced, whose definitions are recalled below.

Fix $\varepsilon>0$ and a Baire Class $1$ function $f:X\to\mathbb{R}.$ All ordinals considered below are countable. Recall that oscillation of $f$ at $x\in X$ on $P\subseteq X$ is defined by $$\omega(f,x,P) = \inf\{ \sup_{y,z\in U\cap P}|f(y)-f(z)| : U\text{ is open}, U\ni x \}.$$ Define iterative derivatives as follows: \begin{align*} D^0(f,\varepsilon,X) & = X \\ D^{\alpha+1}(f,\varepsilon,X) & = \{x\in D^\alpha(f,\varepsilon, X): \omega(f,x, D^\alpha(f,\varepsilon,X)\geq \varepsilon) \} \\ D^\alpha(f,\varepsilon,X) & = \bigcap_{\theta<\alpha}D^\theta(f,\varepsilon,X) \quad \alpha \text{ is a limit ordinal}. \end{align*} One interpretation of the iterative sets is that, we start with the whole space $X.$ Then we collect those discontinuities in $X$ where $f$ 'jump' more than $\varepsilon.$ Among those discontinuities, we do the same for transfinitely many steps.

Let $\beta(f,\varepsilon) = \min\{\eta: D^\eta(f,\varepsilon,X)=\emptyset\}$ and oscillation rank of $f$ be $$\beta(f) = \sup_{\varepsilon>0} \beta(f,\varepsilon).$$

Note that the rank $\beta$ is investigated by Kechris and Denny and Tang.

Question: Does there exist a Baire Class $1$ function $f:X\to\mathbb{R}$ such that $\omega\cdot n<\beta(f)\leq \omega\cdot (n+1)$ for some natural number $n\geq 1.$

We know that $\beta(f)=1$ if $f$ is continuous, $\beta(f)<\omega_1$ (the first uncountable ordinal) if $f$ is Baire Class $1.$ However, I am unable to construct a Baire Class $1$ function whose oscillation rank falls between $\omega$ and $\omega\cdot 2,$ which is $n=1$ in my question. When $n=0,$ I can construct a function $f$ such that $\beta(f)=m$ where $m$ is a given natural number.

• As written, the definition makes $D^\alpha(f,\varepsilon,X)=X$ for all $\alpha$. – Emil Jeřábek Jun 1 '18 at 14:52
• @EmilJeřábek Thanks for spotting the mistake. I have edited my question. – Idonknow Jun 1 '18 at 15:06

What about the function $f:\omega_1\to\{0,1\}$ assigning to each ordinal $\alpha$ written in the Cantor normal form $\alpha=\omega^{\beta_1}c_1+\cdots\omega^{\beta_k}c_k$ the number $f(\alpha):=c_k\mod 2$?
It seems that the restriction of $f$ to a suitably large countable ordinal (which is a locally compact Polish scattered space) can have arbitrarily high oscillation rank $\beta(f)$.
In particular, $f|(\omega^\omega+1)$ has the oscillation rank $\beta(f)=\omega+1$.
Since the space $\omega_1$ is scattered, for every non-empty subset $A\subset \omega_1$ the restriction $f|A$ has a continuity point (which is a bit better than being of the first Baire class).