Let $L(\bar{\boldsymbol{w}})$ is an arbitrary $0/1$ loss function (use in perceptron algorithm). Let $v$ is a value of its range. My question is about the property of the set $L^{-1}(v)$. Is it a connected set?
Definition of 0/1 loss function. Let $D = \{(\boldsymbol{x_1},y_1),...,(\boldsymbol{x_N},y_N)\}$, where $\boldsymbol{x_i}\in \mathbb{R}^n$, $y_i\in \{-1,1\}$. Let $f:=\boldsymbol{w\cdot x}+w_0$, where $\boldsymbol{w,x}\in \mathbb{R}^n, w_0\in \mathbb{R}$, "$\cdot$" is dot product. With $\boldsymbol{w}=(w_1,...,w_n),\boldsymbol{x}=(x_1,...,x_n),$ define $\bar{\boldsymbol{w}}:=(w_0,w_1,...,w_n),\bar{\boldsymbol{x}}:=(x_0=1,x_1,...,x_n).$
Then $f$ can rewrite as $f=\bar{\boldsymbol{w}}\cdot \bar{\boldsymbol{x}}$. The $0/1$ loss function with respect to $D$ is defined as $L(\bar{\boldsymbol{w}}):=$ the number of $(x_i,y_i)\in D$ such that $f(\bar{\boldsymbol{x_i}}).y_i<0$.
N,n positive integers; * dot product; |A| cardinality A
D = ((a1,d1),.. (a_N,d_N)) in (R^n x {-1,1})^N
L:R^n x R -> Z, (x,r) -> |{ i in {1,.. N} : (a_i*x + r).d_i < 0 }|
For k in Z, is L^-1(k) connected? Is L^-1(k) simply connected?
No. D = ((0^n,-1), (0^n,1)) is counter example.
(a_1*x + r)d_1 = -r; (a_2*x + r)d_2 = r
L(x,0) = 0; r /= 0 implies L(x,r) = 1; L^-1(1) = R^n x (R - {0})