Distance between two sets different from zero

Question

Consider these sets $$ A\equiv \bigcap_{\delta>0} \liminf_{n\rightarrow \infty} \{x \in X: d(p_n, [\ell(x), u(x)])\leq \delta\} $$ $$ A_n\equiv \{x \in X: d(p_n, [\ell(x), u(x)])=0\} $$ where:

• $A$ is non-empty.
• $(p_n)_n$ is a sequence of reals taking values in $[0,1]$.
• $\ell(\cdot)$ and $u(\cdot)$ are real function taking values in $(0,1)$ with $\ell<u$.
• $d\big(p_n, [\ell(x), u(x) ] \big):= \inf \big\{|p_n - y| : y \in [\ell(x), u(x) ] \big\}$.

Let $$ d_H(A, B)\equiv \max\{\sup_{x\in B}d(x,A), \sup_{x\in A}d(x, B)\}, $$ denote the Hausdorff distance. Can you give me a simple example of why $d_H(A,A_n)$ may be different from zero?

Answer 1

Take $X=\mathbb{R}$, $\ell,u:\mathbb{R}\to[0,1]$ given by $\ell(x)=u(x)=0$ for all $x\in\mathbb{R}$, and $\{p_j\}_{j\in\mathbb{N}}$ given by

$$p_j=\begin{cases} 1,&j=0,\\ 0,&j\geq1. \end{cases}$$

Then, for any $\delta>0$,

\begin{align*} \liminf_{j\to\infty}\{x\in\mathbb{R}:d(p_j,[\ell(x),u(x)])\leq\delta\} &=\bigcup_{n\geq0}\bigcap_{j\geq n}\{x\in\mathbb{R}:d(p_j,[\ell(x),u(x)])\leq\delta\} \\ &=\{x\in\mathbb{R}:d(0,\{0\})\leq\delta\} \\ &=\{x\in\mathbb{R}:0\leq\delta\} \\ &=\mathbb{R}. \end{align*}

Consequently

$$A=\bigcap_{\delta>0}\liminf_{j\to\infty}\{x\in\mathbb{R}:d(p_j,[\ell(x),u(x)])\leq\delta\}=\bigcap_{\delta>0}\mathbb{R}=\mathbb{R}.$$

However

$$A_0=\{x\in\mathbb{R}:d(p_0,[\ell(x),u(x)])=0\}=\{x\in\mathbb{R}:d(1,\{0\})=0\}=\varnothing.$$

Clearly then

$$d_H(A,A_0)=d_H(\mathbb{R},\varnothing)=\infty.$$

Answer 2

Given a sequence $(S_n)_{n\in\mathbb N}$ of sets, I assume that $$\liminf_{n\rightarrow \infty} S_n=\bigcup_{n\in\mathbb N} \bigcap_{m\ge n} S_m.$$

Let $X$ be the interval $[1/3,2/3]$ endowed with the usual metric $d$. For each $x\in X$ put $\ell(x)=x-1/6$ and $u(x)=x+1/6$. For each $n\in\mathbb N$ let $p_n=1/3$, if $n$ is odd, and $p_n=2/3$, if $n$ is even. Then for each natural $n$ we have $$A_n=\{x \in X: d(p_n, [\ell(x), u(x)])=0\}=$$ $$\{x \in [1/3,2/3]: x-1/6\le p_n\le x+1/6\}.$$ So $A_n=[1/3,1/2]$, if $n$ is odd and $A_n=[1/2,2/3]$, if $n$ is even.

Let $\delta\le 1/6$ be any positive number. Then for each natural $n$ we have $$\{x \in X: d(p_n, [\ell(x), u(x)])\le\delta\}=$$ $$\{x \in [1/3,2/3]: x-1/6-\delta\le p_n\le x+1/6+\delta\}.$$ The latter set is $[1/3,1/2+\delta]$, if $n$ is odd and $[1/2-\delta,2/3]$, if $n$ is even. Then $\liminf_{n\rightarrow \infty} \{x \in X: d(p_n, [\ell(x), u(x)])\leq \delta\}=[1/2-\delta,1/2+\delta]$, so $A=\{1/2\}$, and thus $d_H(A,A_n)=1/6$ for each natural $n$.