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Consider these sets Aδ>0lim inf 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?

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  • \begingroup I guess you wanted \cup instead of \cap everywhere. \endgroup Commented Mar 21 at 16:57
  • \begingroup This about this a bit more. \endgroup Commented Mar 22 at 19:10
  • \begingroup I wanted to say: Think about this a bit more. Sorry for for the typo. \endgroup Commented Mar 24 at 11:46
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    \begingroup It will be true if \cap is replaced by \cup, or if L on the right-hand side is replaced by 0. \endgroup Commented Mar 24 at 21:02
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    \begingroup According to these guidlelines, users should "avoid trying to answer questions which [...] request answers to multiple questions". Therefore, I suggest that the very non-specific "Can we say something" question be removed, leaving only the question "does this distance go to zero?" \endgroup Commented Mar 24 at 23:38

1 Answer 1

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\newcommand\de\deltaThe 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.


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.

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  • \begingroup Thanks. Is your example showing that there is no L_n that makes the claim true? \endgroup
    – Star
    Commented Mar 25 at 17:05
  • \begingroup @Star : Essentially, this example shows that, for any positive sequence (L_n) converging to 0, we can construct X,\ell,u,(p_n) with d_H(A,C_n(L_n)) however large or even infinite for all n. I believe that all this pursuit, without serious structural restrictions on \ell and u (such as a seriously strong version of injectivity) is, unfortunately, fruitless. \endgroup Commented Mar 25 at 17:29
  • \begingroup But it seems that you have picked a specific L_n? \endgroup
    – Star
    Commented Mar 25 at 17:55
  • \begingroup @Star : I did that in order for the counterexample to be simple and specific. However, the counterexample can be easily modified for any positive sequence (L_n) converging to 0 -- by then letting p_n:=2L_n, say. There are a huge number of degrees of freedom in constructing a counterexample under your conditions. \endgroup Commented Mar 25 at 18:16
  • \begingroup If it is not a huge work, could you edit your answer along those lines? \endgroup
    – Star
    Commented Mar 25 at 18:30

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