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Consider $Q = [0,1]^2$ and $\phi:Q\to \mathbb{R}_{\geq 0}$ such that $\phi$ is a Lipschitz continuous function with a constant, say $\lambda_\phi$.

For each $x\in[0,1]$ we put $S(x) = supp \phi(x,\cdot)$ - i.e. $S(x)$ is a closure of the following set $$ (y\in[0,1]:\phi(x,y)>0). $$

We call a non-empty set $A\subseteq [0,1]$ self-complete if for any $x\in A$ holds $S(x)\subseteq A$. Could you give some ideas how to verify if there are self-complete sets? Any ideas are more than welcome.

What I was able to prove is that $A$ is a set of a positive measure and if there is a self-complete set $A$ then its closure is also self-complete.

P.S. The term self-complete I use myself - maybe there are intersection in a terminology.

Edited: the problem can be reformulated as following: given an open set $G\subset Q$ verify if there is a closed non-empty set $A\subset [0,1]$ such that if $x\in A$ and $(x,y)\in \bar{G}$ then $y\in A$.

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    So $A$ is self-complete if and only if $A\supseteq\bigcup_{x\in [0,1]} S(x)$?2011-04-15
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    No, if and only if $\bigcup_{x\in A}S(x)\subseteq A$.2011-04-15
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    Oh, I see. That makes it more non-trivial. ;)2011-04-15
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    The empty set is self-complete and has measure zero.2011-04-15
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    Sorry, I've forgot about this. I've changed the definition to exclude empty sets. Just forgot to write it for the first time.2011-04-15
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    Another cheap solution: $A=[0,1]$ is self-complete. You should say $S(x)$ is the closure of ... as there is only one. Do you want $A$ self-complete for all $\phi$ or for one specific $\phi$?2011-04-15
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    I mildly changed the definition and removed the set-theory tag.2011-04-15
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    To be fair, I'm not 100% sure what you're asking here.2011-04-17
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    I've added a simple reformulation.2011-04-17
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    The answer to your new formulation is "no" if $\subset$ means proper subset. Let $G = Q - \{ (x,0) : x \in [0,1]\}$, which is a proper subset of $Q$ and open as well (its complement, the $x$-axis, is closed). Any nonempty $A$ with your properties will have $(0,1] \subseteq A$ and then, since $A$ is also closed, $[0,1] \subseteq A$. This corresponds with the constant function $\phi(x,y) = 1$, as in my answer below.2011-04-17
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    @Carl Memmer: are you sure? You provide only one example of $G$ when it's not satisfied. For any other $G$ it can be. E.g. $G = (0.4,0.6)^2$ and $A = [0.4,0.6]$. By the way, I edited a little bit formulation, but it doesn't matter for your argument.2011-04-17
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    One counterexample is enough to show that a conjecture is false.2011-04-17
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    I didn't say "for any open set $G$". I said that for a given set I would like to verify if there a self-complete set. Your example considers only one $G$ and shows that there is no $A$. So your comment doesn't give an answer, please, be correct.2011-04-17
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    It does give an answer: in general, such a set $A$ does not exist.2011-04-17

1 Answers 1

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Several different thoughts. I take it that a set $A \subseteq [0,1]$ is self-complete if $\bigcup_{x \in A} S(x) \subseteq A$, where $S \colon [0,1] \to P([0,1])$ is some function defined as above in terms of a single fixed Lipschitz $\phi\colon [0,1]^2 \to \mathbb{R}_{\geq 0}$.

(1) In some cases the set will be $[0,1]$ itself, for example if $\phi(x,y) = 1$ then $[0,1]$ is the only (nonempty) self-complete set.

(2) Regardless what $S$ is, we can make a self-complete set by just creating a set that is "closed under $S$" in an appropriate sense. You simply use transfinite induction to work towards the goal that whenever $x \in A$, $S(x) \subseteq A$. It goes like this:

Pick any point $x_0$ and let $A_0 = \{ x_0\}$. Now, by transfinite induction, for $\lambda > 0$ let

$$A_\lambda = \left ( \bigcup_{\kappa < \lambda} A_\kappa \right ) \cup \left ( \bigcup_{\kappa < \lambda}\, \bigcup_{x \in A_\kappa} S(x) \right ) $$

This has the property that $A_\kappa \subseteq A_\lambda$ whenever $\kappa < \lambda$. We can't keep adding new points forever, because there are only as many points as the cardinality of [0,1]. So eventually we will have $A_\kappa = A_{\kappa+1}$, and this will be a self-complete set.

(3) Normally, you could replace this transfinite induction with a "top-down" argument. In fact the intersection of any family of self-complete sets would be self complete except for the requirement you added that self-complete sets must be nonempty. Without that requirement, each $\phi$ would be associated with a particular minimal self-complete set. The argument in part (2) above shows that for any $x \in [0,1]$ there is a nonempty self-complete set containing $x$. The intersection of all such sets will still be a self-complete set containing $x$, and so will be a minimal self-complete set among the ones that contain $x$. In fact, this is the set that was constructed in part (2).

(4) You cannot prove that every self-complete set is of positive measure. If $\phi$ is identically 0 then $S(x) = \varnothing$ for every $x$, and so every (nonempty) set is self-complete.

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    To slightly pick on the transfinite argument, I believe that you can limit yourself to $\omega_1$ by arguments of density (as you require $S$ to return a closed set).2011-04-17
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    I understand the general sense of what you mean, but it doesn't seem obvious to me how to prove it. If there is an easy proof of it, could you write it here? In the proof above I just stuck to the general principle and didn't think at all about when the induction would end; all that's needed is some fixed point.2011-04-17
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    I will give it some thought, anything interesting I will post in the comments.2011-04-17
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    (3): "The argument in part (2) above shows that..." isn't it obvious - $[0,1]$ is always self-complete. On the other hand, if $\phi \equiv 1$ then there are no other self-complete sets.2011-04-17
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    But for any $\phi$ there is some *minimal* self-complete set, which will not in general be all of $[0,1]$, although it might be empty. If you take $\phi$ to be supported on the upper half of the square, the minimal nonempty self-complete set will be $[0.5,1]$.2011-04-17