1. **Stating the problem:** We are given sets $A$, $B$, $S_0$, $S_1$, and $S_r$ with the relations $A \subset S_0$, $S_1 = X \setminus B$, and the chain of inclusions $$\overline{A} \subset \overline{S_0} \subset S_r \subset S_1 \subset X \setminus B$$ for all $r \in (0,1)$. We want to understand the definition of the function $f: X \to [0,1]$ given by these sets. 2. **Understanding the sets and inclusions:** - $A \subset S_0$ means $A$ is contained in $S_0$. - $S_1 = X \setminus B$ means $S_1$ is the complement of $B$ in $X$. - The chain $$\overline{A} \subset \overline{S_0} \subset S_r \subset S_1 \subset X \setminus B$$ shows nested sets with closures and intermediate sets $S_r$ for $r \in (0,1)$. 3. **Defining the function $f(x)$:** The function $f$ is defined on $X$ with values in $[0,1]$ by $$ f(x) = \inf \{ r \in (0,1) : x \in S_r \} $$ This means for each $x \in X$, $f(x)$ is the smallest $r$ in $(0,1)$ such that $x$ belongs to $S_r$. 4. **Explanation:** - Since $S_r$ are nested sets increasing with $r$, $f(x)$ measures how "deep" $x$ is inside these sets. - If $x$ is in $\overline{A}$, then $f(x) = 0$ (or close to 0) because $x$ is in all $S_r$ for small $r$. - If $x$ is outside $S_1$, then $f(x) > 1$ or undefined, but since $S_1 = X \setminus B$, $f$ maps into $[0,1]$. 5. **Summary:** The function $f$ assigns to each point $x$ the minimal parameter $r$ for which $x$ lies in the set $S_r$, capturing the nested structure of these sets between $\overline{A}$ and $X \setminus B$. **Final answer:** $$ f(x) = \inf \{ r \in (0,1) : x \in S_r \} $$