Grayscale Functions

Let $X$ be a topological space. For $A⊆X$, let $ι_A(x)=1$ if $x∈A$, and $ι_A(x)=0$ otherwise. Let $\mathcal{U}(X)=\{U~|~U⊆X \text{ is open}\}$. When $(U_n)$ is a sequence in $\mathcal{U}(X)$, we write $U_n→U$ for $U∈\mathcal{U}(X)$ if $$\overline{U}=\overline{\bigcup_{n∈ℕ}(\bigcap_{N≥n} U_N)}$$ (Convergence need not be unique) If $φ : X→\mathcal{U}(X)$ is such that $x∈φ(x)$ when $φ(x)≠ø$, and $φ(x_n)→φ(x)$ whenever $x_n→x$, we say $φ$ is a gradated open assignment on $X$. Given a sequence $σ=(x_i)$ in $X$ and a subset $A⊆X$, let $$λ_σ(A)=\lim_{N→∞} \sum^N \dfrac{ι_A(x_i)}{N}$$ wherever the convergence exists. If $φ$ is a gradated open assignment such that $f = λ_σ ∘ φ : X→[0,1]$ is a well-defined continuous mapping, then the pair $(φ,σ)$ is called a grayscale pair and $f$ the corresponding grayscale function. If every continuous function $f∈\mathcal{C}(X,[0,1])$ is a grayscale function, then $X$ is said to be a grayscale domain.
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Every infinite discrete space is grayscale, and a finite discrete space is never grayscale. In fact, if $X=\{x_1,...,x_n\}$ is a finite discrete space, then the function $f : X→[0,1]$ is grayscale if and only if $f(x_i)=0,1$ for some $i$, or there exists some $n×n$ binary matrix $M$ with $1$s on the main diagonal and a non-negative-valued $n×1$ column vector $A$ whose entries sum to $1$ such that $$MA=\begin{bmatrix} f(x_1) \\ f(x_2) \\ \cdots \\ f(x_n) \end{bmatrix}$$ ~~~~~ We prove $[0,1]$ is itself grayscale. If $f : [0,1]→[0,1]$ is continuous, then let $$φ(x) = \left\{ \begin{array} \{ [0,f(x)) & x=0 \\ (1-f(x),1] & x=1 \\ (x(1-f(x)),~f(x)+x(1-f(x)) & x∈(0,1) \end{array} \right.$$ Note that the length of the interval $φ(x)$ is $f(x)$, and $x∈φ(x)$ if $φ(x)≠ø$. To prove gradation, note that when $x_n→x$, we have $x_n(1-f(x_n))→x(1-f(x))$ and $f(x_n)+x_n(1-f(x_n))→f(x)+x(1-x)$. When $μ(n)→0$ denotes the maximum of the difference between the two pairs of an $n^\text{th}$ term and its limit, we see that if $B(y,ε)⊆φ(x)$ for $x∈(0,1)$, we have $y∈φ(x_n)$ for all $n≥N$ when $N$ is such that $μ(n) < ε$ for all $n≥N$. Exclusion of those elements $y∉\overline{φ(x)}$ from the final intersection is similar. The cases for $x=0,1$ are straightforward edge cases.

Let $σ=(x_n)$ be the sequence of rationals in the order $\dfrac{1}{2},\dfrac{1}{3},\dfrac{2}{3},\dfrac{1}{4},\dfrac{2}{4},\dfrac{3}{4},...$ (we shall refer to the subcollection of those with $n$ in the denominator as the $n$-strip elements).

It now remains to show $λ_σ(x)=f(x)$. To observe this, first fix $x$ and let $α=f(x)$. Note that each $n$-strip collection refers to a subset of $[0,1]$ containing $n-1$ elements spaced $1/n$ apart from each other. Therefore, when $g(n)$ denotes the number of $n$-strip elements contained in $φ(x)$, we see $g(n)∈[α(n-1),~α(n-1)-2]$, and when $$G(n)=\sum^N_{i=1} g(i)$$ we see $G(n)∈[\dfrac{1}{2}αn(n-1)-2n,~\dfrac{1}{2}αn(n-1)]$. Since the number of elements up to and including the $n$-strip elements is $β(n)=\dfrac{1}{2}n(n-1)$, we see $G(n)/\sum β(n)∈[α-\dfrac{4}{n-1},α]$. This shows that a subsequence of the limit involved in $λ_σ(x)$ converges to $α=f(x)$, so it suffices to show that the limit inferior and limit superior are the same. To wit, note that the minimum value between the partial sums $G(n)/\sum β(n)$ and $G(n+1)/\sum β(n+1)$ is bounded below by $$G(n)/\sum β(n+1)∈[α\dfrac{n-5}{n+1},α\dfrac{n-1}{n+1}]$$ The case for limit superior is parallel. Thus $λ_σ(x)$ is well defined and equals $f(x)$.

By a similar argument, $(0,1)≅ℝ$ is also grayscale.