Convergence of distance

Question

Consider these sets $$ A\equiv \bigcap_{\delta>0} \liminf_{n\rightarrow \infty} \{x \in X: d(p_n, [\ell(x), u(x)])\leq \delta\} $$ $$ C_n(L_n)\equiv \{x \in X: d(p_n, [\ell(x), u(x)])=0\} $$ 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 $[0,1]$.
• $d\big(p_n, [\ell(x), u(x) ] \big):= \inf \big\{|p_n - y| : y \in [\ell(x), u(x) ] \big\}$.

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. Is $d_H(A,C_n(L_n))= 0$?

Answer 1

$\newcommand\de\delta$The answer is no. In fact, $d_H(A,C_n(L_n))$ can be however large or even infinite for all $n$.

Indeed, e.g. suppose that $X=[-1,a)$ for some $a\in(0,\infty]$ endowed with the standard metric $|\cdot-\cdot|$, $\ell(x)=u(x)=-x$ for $x\in[-1,0]$, $\ell(x)=u(x)=0$ for $x\in[0,a)$. Let $(L_n)$ be any positive sequence converging to $0$, and let $p_n=2L_n$ eventually (that is, for all large enough $n$). Then $A\supseteq[0,a)$ and $C_n(L_n)=[-3L_n,-L_n]$ eventually, so that eventually $$d_H(A,C_n(L_n))\ge|(a-L_n)-(-L_n)|=a$$ if $a<\infty$, and $d_H(A,C_n(L_n))=\infty$ if $a=\infty$.

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Details: Let $A_n(\de):=\{x\in X\colon d(p_n,[\ell(x),u(x)])\le\de\}$ and $A(\de):=\liminf_{n\to\infty}A_n(\de)$, so that $A=\bigcap_{\de>0}A(\de)$.

For each $\de>0$, eventually $A_n(\de)\supseteq[0,a)$, so that $A(\de)\supseteq[0,a)$ for all $\de>0$ and hence $A\supseteq[0,a)$, as was claimed.

Also, $C_n(L_n)=A_n(L_n)=\{x\in[-1,0]\colon |p_n-(-x)|\le L_n\}=[-3L_n,-L_n]$ eventually, as was claimed, too.