# Binary classification problem

This problem is in the context of binary classification. Let $$f_\omega (t) = \mathcal I[\sin (\omega t) \geq 0]$$, and $$\mathcal F = \{ f_{\omega} : \omega \in R\}$$, $$t\in\mathbb R$$. For any given $$m = 1, 2, 3, \dots$$ find a set of $$m$$ points $$x_1, x_2, \dots, x_m$$ on $$\mathbb R$$ such that for every point $$x_i$$ and an arbitrary "label" $$y_i \in \{ 0, 1\}$$ associated to it, we can always find $$g\in\mathcal F$$ such that $$g(x_i) = y_i$$ for every $$i \in [1, m]$$. I have to use the following fact. For any $$z\in (0, 1)$$, $$\mathcal I[\sin(2^i \pi z) \geq 0] = 1 - z_i$$, where $$z_i$$ is the $$i$$-th "bit" of the base-2 representation of $$z$$, e.g., $$z = \sum^{\infty}_{i=1}z_i 2^{-i}$$.

I'm stuck at how to choose the $$m$$ points. My initial guess was to set $$x_i = 2^{-i}$$, but I don't see how can I use the fact I have to use since the labels should be arbitrary. Any hint?

Define $$p_n(t)=f_{2^n\pi}(t)=[\sin2^n\pi t\ge0]$$ and let $$x_1=\frac23=0.\overline{10}_2$$. Then the given fact implies that $$p_k(x_1)$$ is the last bit of $$k$$. Similarly we choose $$x_2=0.\overline{1100}_2=\frac45,x_3=0.\overline{11110000}_2=\frac{16}{17}$$, etc. – in general $$x_n=0.\overline{\underbrace{1\dots1}_{2^{n-1}}\underbrace{0\dots0}_{2^{n-1}}}_2=1-\frac1{2^{2^{n-1}}+1}$$ and $$p_k(x_n)$$ is the $$n$$th least significant bit of $$k$$ for nonnegative integer $$k$$.
With these $$x_i$$, if we are given arbitrary labels $$y_i$$ forming the integer $$z=(y_m\dots y_1)_2$$, we simply choose $$g=p_z=f_{2^z\pi}$$. For $$1\le i\le m$$, $$g(x_i)=p_z(x_i)$$ will give the $$i$$-th least significant bit of $$z$$, i.e. $$y_i$$.